more efficient way than using a 'for' loop












1















I feel there's smarter/more efficient way than this code:



df <- mtcars

df$somename <- as.array(rep(c(0), 32))

for (i in 1:32){
df$somename[i] <- sd(c(df$wt[i], df$qsec[i]))
}


maybe with %>%? but how?










share|improve this question


















  • 1





    This is actually a loop as well but one alternative is mapply(function(x, y) sd(c(x, y)), df$wt, df$qsec) OR sapply(1:nrow(df), function(x) sd(c(df$wt[x], df$qsec[x])))

    – Ronak Shah
    Nov 26 '18 at 5:20
















1















I feel there's smarter/more efficient way than this code:



df <- mtcars

df$somename <- as.array(rep(c(0), 32))

for (i in 1:32){
df$somename[i] <- sd(c(df$wt[i], df$qsec[i]))
}


maybe with %>%? but how?










share|improve this question


















  • 1





    This is actually a loop as well but one alternative is mapply(function(x, y) sd(c(x, y)), df$wt, df$qsec) OR sapply(1:nrow(df), function(x) sd(c(df$wt[x], df$qsec[x])))

    – Ronak Shah
    Nov 26 '18 at 5:20














1












1








1








I feel there's smarter/more efficient way than this code:



df <- mtcars

df$somename <- as.array(rep(c(0), 32))

for (i in 1:32){
df$somename[i] <- sd(c(df$wt[i], df$qsec[i]))
}


maybe with %>%? but how?










share|improve this question














I feel there's smarter/more efficient way than this code:



df <- mtcars

df$somename <- as.array(rep(c(0), 32))

for (i in 1:32){
df$somename[i] <- sd(c(df$wt[i], df$qsec[i]))
}


maybe with %>%? but how?







r






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 26 '18 at 5:02









Tony DTony D

1168




1168








  • 1





    This is actually a loop as well but one alternative is mapply(function(x, y) sd(c(x, y)), df$wt, df$qsec) OR sapply(1:nrow(df), function(x) sd(c(df$wt[x], df$qsec[x])))

    – Ronak Shah
    Nov 26 '18 at 5:20














  • 1





    This is actually a loop as well but one alternative is mapply(function(x, y) sd(c(x, y)), df$wt, df$qsec) OR sapply(1:nrow(df), function(x) sd(c(df$wt[x], df$qsec[x])))

    – Ronak Shah
    Nov 26 '18 at 5:20








1




1





This is actually a loop as well but one alternative is mapply(function(x, y) sd(c(x, y)), df$wt, df$qsec) OR sapply(1:nrow(df), function(x) sd(c(df$wt[x], df$qsec[x])))

– Ronak Shah
Nov 26 '18 at 5:20





This is actually a loop as well but one alternative is mapply(function(x, y) sd(c(x, y)), df$wt, df$qsec) OR sapply(1:nrow(df), function(x) sd(c(df$wt[x], df$qsec[x])))

– Ronak Shah
Nov 26 '18 at 5:20












3 Answers
3






active

oldest

votes


















4














An option using purrr::map2



library(tidyverse)
mtcars %>% mutate(somename = map2(wt, qsec, ~sd(c(.x, .y))))
# mpg cyl disp hp drat wt qsec vs am gear carb somename
#1 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4 9.786358
#2 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4 10.00203
#3 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1 11.51877
#4 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1 11.47281
#5 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2 9.60251
#6 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1 11.85111
#7 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4 8.6762
#8 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2 11.88646
#9 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2 13.96536
#10 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4 10.50761
#11 17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4 10.93187
#12 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3 9.425733
#13 17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3 9.807571
#14 15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3 10.05506
#15 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4 9.001469
#16 10.4 8 460.0 215 3.00 5.424 17.82 0 0 3 4 8.765296
#17 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4 8.538314
#18 32.4 4 78.7 66 4.08 2.200 19.47 1 1 4 1 12.21173
#19 30.4 4 75.7 52 4.93 1.615 18.52 1 1 4 2 11.95364
#20 33.9 4 71.1 65 4.22 1.835 19.90 1 1 4 1 12.77388
#21 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1 12.40619
#22 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2 9.439876
#23 15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2 9.804036
#24 13.3 8 350.0 245 3.73 3.840 15.41 0 0 3 4 8.181225
#25 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2 9.337345
#26 27.3 4 79.0 66 4.08 1.935 18.90 1 1 4 1 11.99607
#27 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2 10.29547
#28 30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2 10.88025
#29 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4 8.01152
#30 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6 9.001469
#31 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8 7.799388
#32 21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2 11.18643




Update



I re-ran @42-'s microbenchmark analysis using a larger dataset



library(microbenchmark)
df <- do.call(rbind, lapply(1:100, function(x) mtcars))
res <- microbenchmark(
orig = {
df$somename <- as.array(rep(c(0), nrow(df)))
for (i in 1:nrow(df)) {
df$somename[i] <- sd(c(df$wt[i], df$qsec[i]))}},
tidy = {
df <- df %>% mutate(somename = map2(wt, qsec, ~sd(c(.x, .y))))},
mapply = {
df$somename <- mapply(function(x, y) sd(c(x, y)), df$wt, df$qsec)},
rowMeans = {
df$rm <- rowMeans(df[,c("wt","qsec")])
df$sd2col <- sqrt( (df$wt - df$rm)^2 + (df$qsec - df$rm)^2 )})
res
#Unit: microseconds
# expr min lq mean median uq max
# orig 331092.86 349754.808 360716.6501 357229.3920 366635.2820 446581.924
# tidy 168701.28 181079.910 189710.1927 187026.6290 194392.5190 273725.354
# mapply 161711.77 172457.395 179326.5484 177263.3045 183688.5365 266102.901
# rowMeans 228.08 315.854 343.9151 334.8975 358.5915 807.847

library(ggplot2)
autoplot(res)


enter image description here






share|improve this answer


























  • The mapply solution is faster.

    – 42-
    Nov 26 '18 at 5:33











  • @42- I agree. But part of OPs question seems to ask for a tidyverse-eque approach.

    – Maurits Evers
    Nov 26 '18 at 5:33











  • @chinsoon12 Aaah bugger!! You're absolutely right. Let me re-run...

    – Maurits Evers
    Nov 26 '18 at 6:35













  • @chinsoon12 ... and done. Ok, so mapply and tidy scale very similarly, with mapply being still slightly faster.

    – Maurits Evers
    Nov 26 '18 at 6:40






  • 1





    for this particular specific example, df$rm <- rowMeans(df[,c("wt","qsec")]) ; df$sd2col <- sqrt( (df$wt - df$rm)^2 + (df$qsec - df$rm)^2 ) would be faster

    – chinsoon12
    Nov 26 '18 at 6:42



















2














More of a comment than an answer:



> library(microbenchmark)
> microbenchmark( orig = {df <- mtcars
+
+ df$somename <- as.array(rep(c(0), 32))
+
+ for (i in 1:32){
+ df$somename[i] <- sd(c(df$wt[i], df$qsec[i]))
+ }}, tidy = {
+ mtcars %>% mutate(somename = map2(wt, qsec, ~sd(c(.x, .y))))}, mapply = { mapply(function(x, y) sd(c(x, y)), df$wt, df$qsec)})

#------------------------------------
Unit: microseconds
expr min lq mean median uq max neval cld
orig 5069.391 5161.9270 5555.5886 5236.769 5490.7365 12400.502 100 b
tidy 910.071 943.9685 986.4419 970.541 998.8075 1241.711 100 a
mapply 744.639 761.1875 805.6328 773.426 807.2545 2206.393 100 a





share|improve this answer
























  • Just out of curiosity, I re-ran the benchmark analysis using a larger dataset (see my updated post), and the for loop solution ended up being significantly faster than both the mapply and tidyverse approach. Not much difference between mapply and purrr::map2.

    – Maurits Evers
    Nov 26 '18 at 6:22













  • Ah I made a blunder of things (definitely time to call it a day;-). It's fixed now. mapply and tidy scale very similarly with larger datasets (and are faster than orig), with mapply being slightly faster than tidy.

    – Maurits Evers
    Nov 26 '18 at 6:42



















1














Code:



df$somename <- apply(matrix(c(df$wt, df$qsec), ncol=2), MARGIN = 1, FUN=sd)



Output:



> head(df$somename)

somename
1 9.786358
2 10.002025
3 11.518769
4 11.472808
5 9.602510
6 11.851110
7 8.676200
8 11.886465
9 13.965359
10 10.507607





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4














    An option using purrr::map2



    library(tidyverse)
    mtcars %>% mutate(somename = map2(wt, qsec, ~sd(c(.x, .y))))
    # mpg cyl disp hp drat wt qsec vs am gear carb somename
    #1 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4 9.786358
    #2 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4 10.00203
    #3 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1 11.51877
    #4 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1 11.47281
    #5 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2 9.60251
    #6 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1 11.85111
    #7 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4 8.6762
    #8 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2 11.88646
    #9 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2 13.96536
    #10 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4 10.50761
    #11 17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4 10.93187
    #12 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3 9.425733
    #13 17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3 9.807571
    #14 15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3 10.05506
    #15 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4 9.001469
    #16 10.4 8 460.0 215 3.00 5.424 17.82 0 0 3 4 8.765296
    #17 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4 8.538314
    #18 32.4 4 78.7 66 4.08 2.200 19.47 1 1 4 1 12.21173
    #19 30.4 4 75.7 52 4.93 1.615 18.52 1 1 4 2 11.95364
    #20 33.9 4 71.1 65 4.22 1.835 19.90 1 1 4 1 12.77388
    #21 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1 12.40619
    #22 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2 9.439876
    #23 15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2 9.804036
    #24 13.3 8 350.0 245 3.73 3.840 15.41 0 0 3 4 8.181225
    #25 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2 9.337345
    #26 27.3 4 79.0 66 4.08 1.935 18.90 1 1 4 1 11.99607
    #27 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2 10.29547
    #28 30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2 10.88025
    #29 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4 8.01152
    #30 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6 9.001469
    #31 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8 7.799388
    #32 21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2 11.18643




    Update



    I re-ran @42-'s microbenchmark analysis using a larger dataset



    library(microbenchmark)
    df <- do.call(rbind, lapply(1:100, function(x) mtcars))
    res <- microbenchmark(
    orig = {
    df$somename <- as.array(rep(c(0), nrow(df)))
    for (i in 1:nrow(df)) {
    df$somename[i] <- sd(c(df$wt[i], df$qsec[i]))}},
    tidy = {
    df <- df %>% mutate(somename = map2(wt, qsec, ~sd(c(.x, .y))))},
    mapply = {
    df$somename <- mapply(function(x, y) sd(c(x, y)), df$wt, df$qsec)},
    rowMeans = {
    df$rm <- rowMeans(df[,c("wt","qsec")])
    df$sd2col <- sqrt( (df$wt - df$rm)^2 + (df$qsec - df$rm)^2 )})
    res
    #Unit: microseconds
    # expr min lq mean median uq max
    # orig 331092.86 349754.808 360716.6501 357229.3920 366635.2820 446581.924
    # tidy 168701.28 181079.910 189710.1927 187026.6290 194392.5190 273725.354
    # mapply 161711.77 172457.395 179326.5484 177263.3045 183688.5365 266102.901
    # rowMeans 228.08 315.854 343.9151 334.8975 358.5915 807.847

    library(ggplot2)
    autoplot(res)


    enter image description here






    share|improve this answer


























    • The mapply solution is faster.

      – 42-
      Nov 26 '18 at 5:33











    • @42- I agree. But part of OPs question seems to ask for a tidyverse-eque approach.

      – Maurits Evers
      Nov 26 '18 at 5:33











    • @chinsoon12 Aaah bugger!! You're absolutely right. Let me re-run...

      – Maurits Evers
      Nov 26 '18 at 6:35













    • @chinsoon12 ... and done. Ok, so mapply and tidy scale very similarly, with mapply being still slightly faster.

      – Maurits Evers
      Nov 26 '18 at 6:40






    • 1





      for this particular specific example, df$rm <- rowMeans(df[,c("wt","qsec")]) ; df$sd2col <- sqrt( (df$wt - df$rm)^2 + (df$qsec - df$rm)^2 ) would be faster

      – chinsoon12
      Nov 26 '18 at 6:42
















    4














    An option using purrr::map2



    library(tidyverse)
    mtcars %>% mutate(somename = map2(wt, qsec, ~sd(c(.x, .y))))
    # mpg cyl disp hp drat wt qsec vs am gear carb somename
    #1 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4 9.786358
    #2 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4 10.00203
    #3 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1 11.51877
    #4 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1 11.47281
    #5 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2 9.60251
    #6 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1 11.85111
    #7 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4 8.6762
    #8 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2 11.88646
    #9 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2 13.96536
    #10 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4 10.50761
    #11 17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4 10.93187
    #12 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3 9.425733
    #13 17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3 9.807571
    #14 15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3 10.05506
    #15 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4 9.001469
    #16 10.4 8 460.0 215 3.00 5.424 17.82 0 0 3 4 8.765296
    #17 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4 8.538314
    #18 32.4 4 78.7 66 4.08 2.200 19.47 1 1 4 1 12.21173
    #19 30.4 4 75.7 52 4.93 1.615 18.52 1 1 4 2 11.95364
    #20 33.9 4 71.1 65 4.22 1.835 19.90 1 1 4 1 12.77388
    #21 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1 12.40619
    #22 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2 9.439876
    #23 15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2 9.804036
    #24 13.3 8 350.0 245 3.73 3.840 15.41 0 0 3 4 8.181225
    #25 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2 9.337345
    #26 27.3 4 79.0 66 4.08 1.935 18.90 1 1 4 1 11.99607
    #27 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2 10.29547
    #28 30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2 10.88025
    #29 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4 8.01152
    #30 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6 9.001469
    #31 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8 7.799388
    #32 21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2 11.18643




    Update



    I re-ran @42-'s microbenchmark analysis using a larger dataset



    library(microbenchmark)
    df <- do.call(rbind, lapply(1:100, function(x) mtcars))
    res <- microbenchmark(
    orig = {
    df$somename <- as.array(rep(c(0), nrow(df)))
    for (i in 1:nrow(df)) {
    df$somename[i] <- sd(c(df$wt[i], df$qsec[i]))}},
    tidy = {
    df <- df %>% mutate(somename = map2(wt, qsec, ~sd(c(.x, .y))))},
    mapply = {
    df$somename <- mapply(function(x, y) sd(c(x, y)), df$wt, df$qsec)},
    rowMeans = {
    df$rm <- rowMeans(df[,c("wt","qsec")])
    df$sd2col <- sqrt( (df$wt - df$rm)^2 + (df$qsec - df$rm)^2 )})
    res
    #Unit: microseconds
    # expr min lq mean median uq max
    # orig 331092.86 349754.808 360716.6501 357229.3920 366635.2820 446581.924
    # tidy 168701.28 181079.910 189710.1927 187026.6290 194392.5190 273725.354
    # mapply 161711.77 172457.395 179326.5484 177263.3045 183688.5365 266102.901
    # rowMeans 228.08 315.854 343.9151 334.8975 358.5915 807.847

    library(ggplot2)
    autoplot(res)


    enter image description here






    share|improve this answer


























    • The mapply solution is faster.

      – 42-
      Nov 26 '18 at 5:33











    • @42- I agree. But part of OPs question seems to ask for a tidyverse-eque approach.

      – Maurits Evers
      Nov 26 '18 at 5:33











    • @chinsoon12 Aaah bugger!! You're absolutely right. Let me re-run...

      – Maurits Evers
      Nov 26 '18 at 6:35













    • @chinsoon12 ... and done. Ok, so mapply and tidy scale very similarly, with mapply being still slightly faster.

      – Maurits Evers
      Nov 26 '18 at 6:40






    • 1





      for this particular specific example, df$rm <- rowMeans(df[,c("wt","qsec")]) ; df$sd2col <- sqrt( (df$wt - df$rm)^2 + (df$qsec - df$rm)^2 ) would be faster

      – chinsoon12
      Nov 26 '18 at 6:42














    4












    4








    4







    An option using purrr::map2



    library(tidyverse)
    mtcars %>% mutate(somename = map2(wt, qsec, ~sd(c(.x, .y))))
    # mpg cyl disp hp drat wt qsec vs am gear carb somename
    #1 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4 9.786358
    #2 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4 10.00203
    #3 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1 11.51877
    #4 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1 11.47281
    #5 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2 9.60251
    #6 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1 11.85111
    #7 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4 8.6762
    #8 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2 11.88646
    #9 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2 13.96536
    #10 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4 10.50761
    #11 17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4 10.93187
    #12 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3 9.425733
    #13 17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3 9.807571
    #14 15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3 10.05506
    #15 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4 9.001469
    #16 10.4 8 460.0 215 3.00 5.424 17.82 0 0 3 4 8.765296
    #17 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4 8.538314
    #18 32.4 4 78.7 66 4.08 2.200 19.47 1 1 4 1 12.21173
    #19 30.4 4 75.7 52 4.93 1.615 18.52 1 1 4 2 11.95364
    #20 33.9 4 71.1 65 4.22 1.835 19.90 1 1 4 1 12.77388
    #21 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1 12.40619
    #22 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2 9.439876
    #23 15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2 9.804036
    #24 13.3 8 350.0 245 3.73 3.840 15.41 0 0 3 4 8.181225
    #25 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2 9.337345
    #26 27.3 4 79.0 66 4.08 1.935 18.90 1 1 4 1 11.99607
    #27 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2 10.29547
    #28 30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2 10.88025
    #29 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4 8.01152
    #30 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6 9.001469
    #31 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8 7.799388
    #32 21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2 11.18643




    Update



    I re-ran @42-'s microbenchmark analysis using a larger dataset



    library(microbenchmark)
    df <- do.call(rbind, lapply(1:100, function(x) mtcars))
    res <- microbenchmark(
    orig = {
    df$somename <- as.array(rep(c(0), nrow(df)))
    for (i in 1:nrow(df)) {
    df$somename[i] <- sd(c(df$wt[i], df$qsec[i]))}},
    tidy = {
    df <- df %>% mutate(somename = map2(wt, qsec, ~sd(c(.x, .y))))},
    mapply = {
    df$somename <- mapply(function(x, y) sd(c(x, y)), df$wt, df$qsec)},
    rowMeans = {
    df$rm <- rowMeans(df[,c("wt","qsec")])
    df$sd2col <- sqrt( (df$wt - df$rm)^2 + (df$qsec - df$rm)^2 )})
    res
    #Unit: microseconds
    # expr min lq mean median uq max
    # orig 331092.86 349754.808 360716.6501 357229.3920 366635.2820 446581.924
    # tidy 168701.28 181079.910 189710.1927 187026.6290 194392.5190 273725.354
    # mapply 161711.77 172457.395 179326.5484 177263.3045 183688.5365 266102.901
    # rowMeans 228.08 315.854 343.9151 334.8975 358.5915 807.847

    library(ggplot2)
    autoplot(res)


    enter image description here






    share|improve this answer















    An option using purrr::map2



    library(tidyverse)
    mtcars %>% mutate(somename = map2(wt, qsec, ~sd(c(.x, .y))))
    # mpg cyl disp hp drat wt qsec vs am gear carb somename
    #1 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4 9.786358
    #2 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4 10.00203
    #3 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1 11.51877
    #4 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1 11.47281
    #5 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2 9.60251
    #6 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1 11.85111
    #7 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4 8.6762
    #8 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2 11.88646
    #9 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2 13.96536
    #10 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4 10.50761
    #11 17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4 10.93187
    #12 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3 9.425733
    #13 17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3 9.807571
    #14 15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3 10.05506
    #15 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4 9.001469
    #16 10.4 8 460.0 215 3.00 5.424 17.82 0 0 3 4 8.765296
    #17 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4 8.538314
    #18 32.4 4 78.7 66 4.08 2.200 19.47 1 1 4 1 12.21173
    #19 30.4 4 75.7 52 4.93 1.615 18.52 1 1 4 2 11.95364
    #20 33.9 4 71.1 65 4.22 1.835 19.90 1 1 4 1 12.77388
    #21 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1 12.40619
    #22 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2 9.439876
    #23 15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2 9.804036
    #24 13.3 8 350.0 245 3.73 3.840 15.41 0 0 3 4 8.181225
    #25 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2 9.337345
    #26 27.3 4 79.0 66 4.08 1.935 18.90 1 1 4 1 11.99607
    #27 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2 10.29547
    #28 30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2 10.88025
    #29 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4 8.01152
    #30 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6 9.001469
    #31 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8 7.799388
    #32 21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2 11.18643




    Update



    I re-ran @42-'s microbenchmark analysis using a larger dataset



    library(microbenchmark)
    df <- do.call(rbind, lapply(1:100, function(x) mtcars))
    res <- microbenchmark(
    orig = {
    df$somename <- as.array(rep(c(0), nrow(df)))
    for (i in 1:nrow(df)) {
    df$somename[i] <- sd(c(df$wt[i], df$qsec[i]))}},
    tidy = {
    df <- df %>% mutate(somename = map2(wt, qsec, ~sd(c(.x, .y))))},
    mapply = {
    df$somename <- mapply(function(x, y) sd(c(x, y)), df$wt, df$qsec)},
    rowMeans = {
    df$rm <- rowMeans(df[,c("wt","qsec")])
    df$sd2col <- sqrt( (df$wt - df$rm)^2 + (df$qsec - df$rm)^2 )})
    res
    #Unit: microseconds
    # expr min lq mean median uq max
    # orig 331092.86 349754.808 360716.6501 357229.3920 366635.2820 446581.924
    # tidy 168701.28 181079.910 189710.1927 187026.6290 194392.5190 273725.354
    # mapply 161711.77 172457.395 179326.5484 177263.3045 183688.5365 266102.901
    # rowMeans 228.08 315.854 343.9151 334.8975 358.5915 807.847

    library(ggplot2)
    autoplot(res)


    enter image description here







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 26 '18 at 6:48

























    answered Nov 26 '18 at 5:07









    Maurits EversMaurits Evers

    30.1k41636




    30.1k41636













    • The mapply solution is faster.

      – 42-
      Nov 26 '18 at 5:33











    • @42- I agree. But part of OPs question seems to ask for a tidyverse-eque approach.

      – Maurits Evers
      Nov 26 '18 at 5:33











    • @chinsoon12 Aaah bugger!! You're absolutely right. Let me re-run...

      – Maurits Evers
      Nov 26 '18 at 6:35













    • @chinsoon12 ... and done. Ok, so mapply and tidy scale very similarly, with mapply being still slightly faster.

      – Maurits Evers
      Nov 26 '18 at 6:40






    • 1





      for this particular specific example, df$rm <- rowMeans(df[,c("wt","qsec")]) ; df$sd2col <- sqrt( (df$wt - df$rm)^2 + (df$qsec - df$rm)^2 ) would be faster

      – chinsoon12
      Nov 26 '18 at 6:42



















    • The mapply solution is faster.

      – 42-
      Nov 26 '18 at 5:33











    • @42- I agree. But part of OPs question seems to ask for a tidyverse-eque approach.

      – Maurits Evers
      Nov 26 '18 at 5:33











    • @chinsoon12 Aaah bugger!! You're absolutely right. Let me re-run...

      – Maurits Evers
      Nov 26 '18 at 6:35













    • @chinsoon12 ... and done. Ok, so mapply and tidy scale very similarly, with mapply being still slightly faster.

      – Maurits Evers
      Nov 26 '18 at 6:40






    • 1





      for this particular specific example, df$rm <- rowMeans(df[,c("wt","qsec")]) ; df$sd2col <- sqrt( (df$wt - df$rm)^2 + (df$qsec - df$rm)^2 ) would be faster

      – chinsoon12
      Nov 26 '18 at 6:42

















    The mapply solution is faster.

    – 42-
    Nov 26 '18 at 5:33





    The mapply solution is faster.

    – 42-
    Nov 26 '18 at 5:33













    @42- I agree. But part of OPs question seems to ask for a tidyverse-eque approach.

    – Maurits Evers
    Nov 26 '18 at 5:33





    @42- I agree. But part of OPs question seems to ask for a tidyverse-eque approach.

    – Maurits Evers
    Nov 26 '18 at 5:33













    @chinsoon12 Aaah bugger!! You're absolutely right. Let me re-run...

    – Maurits Evers
    Nov 26 '18 at 6:35







    @chinsoon12 Aaah bugger!! You're absolutely right. Let me re-run...

    – Maurits Evers
    Nov 26 '18 at 6:35















    @chinsoon12 ... and done. Ok, so mapply and tidy scale very similarly, with mapply being still slightly faster.

    – Maurits Evers
    Nov 26 '18 at 6:40





    @chinsoon12 ... and done. Ok, so mapply and tidy scale very similarly, with mapply being still slightly faster.

    – Maurits Evers
    Nov 26 '18 at 6:40




    1




    1





    for this particular specific example, df$rm <- rowMeans(df[,c("wt","qsec")]) ; df$sd2col <- sqrt( (df$wt - df$rm)^2 + (df$qsec - df$rm)^2 ) would be faster

    – chinsoon12
    Nov 26 '18 at 6:42





    for this particular specific example, df$rm <- rowMeans(df[,c("wt","qsec")]) ; df$sd2col <- sqrt( (df$wt - df$rm)^2 + (df$qsec - df$rm)^2 ) would be faster

    – chinsoon12
    Nov 26 '18 at 6:42













    2














    More of a comment than an answer:



    > library(microbenchmark)
    > microbenchmark( orig = {df <- mtcars
    +
    + df$somename <- as.array(rep(c(0), 32))
    +
    + for (i in 1:32){
    + df$somename[i] <- sd(c(df$wt[i], df$qsec[i]))
    + }}, tidy = {
    + mtcars %>% mutate(somename = map2(wt, qsec, ~sd(c(.x, .y))))}, mapply = { mapply(function(x, y) sd(c(x, y)), df$wt, df$qsec)})

    #------------------------------------
    Unit: microseconds
    expr min lq mean median uq max neval cld
    orig 5069.391 5161.9270 5555.5886 5236.769 5490.7365 12400.502 100 b
    tidy 910.071 943.9685 986.4419 970.541 998.8075 1241.711 100 a
    mapply 744.639 761.1875 805.6328 773.426 807.2545 2206.393 100 a





    share|improve this answer
























    • Just out of curiosity, I re-ran the benchmark analysis using a larger dataset (see my updated post), and the for loop solution ended up being significantly faster than both the mapply and tidyverse approach. Not much difference between mapply and purrr::map2.

      – Maurits Evers
      Nov 26 '18 at 6:22













    • Ah I made a blunder of things (definitely time to call it a day;-). It's fixed now. mapply and tidy scale very similarly with larger datasets (and are faster than orig), with mapply being slightly faster than tidy.

      – Maurits Evers
      Nov 26 '18 at 6:42
















    2














    More of a comment than an answer:



    > library(microbenchmark)
    > microbenchmark( orig = {df <- mtcars
    +
    + df$somename <- as.array(rep(c(0), 32))
    +
    + for (i in 1:32){
    + df$somename[i] <- sd(c(df$wt[i], df$qsec[i]))
    + }}, tidy = {
    + mtcars %>% mutate(somename = map2(wt, qsec, ~sd(c(.x, .y))))}, mapply = { mapply(function(x, y) sd(c(x, y)), df$wt, df$qsec)})

    #------------------------------------
    Unit: microseconds
    expr min lq mean median uq max neval cld
    orig 5069.391 5161.9270 5555.5886 5236.769 5490.7365 12400.502 100 b
    tidy 910.071 943.9685 986.4419 970.541 998.8075 1241.711 100 a
    mapply 744.639 761.1875 805.6328 773.426 807.2545 2206.393 100 a





    share|improve this answer
























    • Just out of curiosity, I re-ran the benchmark analysis using a larger dataset (see my updated post), and the for loop solution ended up being significantly faster than both the mapply and tidyverse approach. Not much difference between mapply and purrr::map2.

      – Maurits Evers
      Nov 26 '18 at 6:22













    • Ah I made a blunder of things (definitely time to call it a day;-). It's fixed now. mapply and tidy scale very similarly with larger datasets (and are faster than orig), with mapply being slightly faster than tidy.

      – Maurits Evers
      Nov 26 '18 at 6:42














    2












    2








    2







    More of a comment than an answer:



    > library(microbenchmark)
    > microbenchmark( orig = {df <- mtcars
    +
    + df$somename <- as.array(rep(c(0), 32))
    +
    + for (i in 1:32){
    + df$somename[i] <- sd(c(df$wt[i], df$qsec[i]))
    + }}, tidy = {
    + mtcars %>% mutate(somename = map2(wt, qsec, ~sd(c(.x, .y))))}, mapply = { mapply(function(x, y) sd(c(x, y)), df$wt, df$qsec)})

    #------------------------------------
    Unit: microseconds
    expr min lq mean median uq max neval cld
    orig 5069.391 5161.9270 5555.5886 5236.769 5490.7365 12400.502 100 b
    tidy 910.071 943.9685 986.4419 970.541 998.8075 1241.711 100 a
    mapply 744.639 761.1875 805.6328 773.426 807.2545 2206.393 100 a





    share|improve this answer













    More of a comment than an answer:



    > library(microbenchmark)
    > microbenchmark( orig = {df <- mtcars
    +
    + df$somename <- as.array(rep(c(0), 32))
    +
    + for (i in 1:32){
    + df$somename[i] <- sd(c(df$wt[i], df$qsec[i]))
    + }}, tidy = {
    + mtcars %>% mutate(somename = map2(wt, qsec, ~sd(c(.x, .y))))}, mapply = { mapply(function(x, y) sd(c(x, y)), df$wt, df$qsec)})

    #------------------------------------
    Unit: microseconds
    expr min lq mean median uq max neval cld
    orig 5069.391 5161.9270 5555.5886 5236.769 5490.7365 12400.502 100 b
    tidy 910.071 943.9685 986.4419 970.541 998.8075 1241.711 100 a
    mapply 744.639 761.1875 805.6328 773.426 807.2545 2206.393 100 a






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Nov 26 '18 at 5:36









    42-42-

    216k15265402




    216k15265402













    • Just out of curiosity, I re-ran the benchmark analysis using a larger dataset (see my updated post), and the for loop solution ended up being significantly faster than both the mapply and tidyverse approach. Not much difference between mapply and purrr::map2.

      – Maurits Evers
      Nov 26 '18 at 6:22













    • Ah I made a blunder of things (definitely time to call it a day;-). It's fixed now. mapply and tidy scale very similarly with larger datasets (and are faster than orig), with mapply being slightly faster than tidy.

      – Maurits Evers
      Nov 26 '18 at 6:42



















    • Just out of curiosity, I re-ran the benchmark analysis using a larger dataset (see my updated post), and the for loop solution ended up being significantly faster than both the mapply and tidyverse approach. Not much difference between mapply and purrr::map2.

      – Maurits Evers
      Nov 26 '18 at 6:22













    • Ah I made a blunder of things (definitely time to call it a day;-). It's fixed now. mapply and tidy scale very similarly with larger datasets (and are faster than orig), with mapply being slightly faster than tidy.

      – Maurits Evers
      Nov 26 '18 at 6:42

















    Just out of curiosity, I re-ran the benchmark analysis using a larger dataset (see my updated post), and the for loop solution ended up being significantly faster than both the mapply and tidyverse approach. Not much difference between mapply and purrr::map2.

    – Maurits Evers
    Nov 26 '18 at 6:22







    Just out of curiosity, I re-ran the benchmark analysis using a larger dataset (see my updated post), and the for loop solution ended up being significantly faster than both the mapply and tidyverse approach. Not much difference between mapply and purrr::map2.

    – Maurits Evers
    Nov 26 '18 at 6:22















    Ah I made a blunder of things (definitely time to call it a day;-). It's fixed now. mapply and tidy scale very similarly with larger datasets (and are faster than orig), with mapply being slightly faster than tidy.

    – Maurits Evers
    Nov 26 '18 at 6:42





    Ah I made a blunder of things (definitely time to call it a day;-). It's fixed now. mapply and tidy scale very similarly with larger datasets (and are faster than orig), with mapply being slightly faster than tidy.

    – Maurits Evers
    Nov 26 '18 at 6:42











    1














    Code:



    df$somename <- apply(matrix(c(df$wt, df$qsec), ncol=2), MARGIN = 1, FUN=sd)



    Output:



    > head(df$somename)

    somename
    1 9.786358
    2 10.002025
    3 11.518769
    4 11.472808
    5 9.602510
    6 11.851110
    7 8.676200
    8 11.886465
    9 13.965359
    10 10.507607





    share|improve this answer




























      1














      Code:



      df$somename <- apply(matrix(c(df$wt, df$qsec), ncol=2), MARGIN = 1, FUN=sd)



      Output:



      > head(df$somename)

      somename
      1 9.786358
      2 10.002025
      3 11.518769
      4 11.472808
      5 9.602510
      6 11.851110
      7 8.676200
      8 11.886465
      9 13.965359
      10 10.507607





      share|improve this answer


























        1












        1








        1







        Code:



        df$somename <- apply(matrix(c(df$wt, df$qsec), ncol=2), MARGIN = 1, FUN=sd)



        Output:



        > head(df$somename)

        somename
        1 9.786358
        2 10.002025
        3 11.518769
        4 11.472808
        5 9.602510
        6 11.851110
        7 8.676200
        8 11.886465
        9 13.965359
        10 10.507607





        share|improve this answer













        Code:



        df$somename <- apply(matrix(c(df$wt, df$qsec), ncol=2), MARGIN = 1, FUN=sd)



        Output:



        > head(df$somename)

        somename
        1 9.786358
        2 10.002025
        3 11.518769
        4 11.472808
        5 9.602510
        6 11.851110
        7 8.676200
        8 11.886465
        9 13.965359
        10 10.507607






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 26 '18 at 7:27









        Farah NazifaFarah Nazifa

        587512




        587512






























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