How to Check if a Date is Within a List of Intervals in R?
I have two data frames (tibbles) with 2 variables each:
- df.POS: ID (ID variable); DATE (Date of positive lab test)
- df.NEG: ID (ID variable); data (Date of negative lab tests (more than 1 test).
Please note that data is a list variable, created with the nest() function of the tidyr package.
library(tidyverse)
library(lubridate)
# negative tests
dates.neg <- ymd(c('2018-02-01', '2018-02-06', '2018-02-10',
'2018-02-21', '2018-04-05'))
df.NEG <- tibble(ID = paste0('ID_', rep(1, 5)),
DATE = dates.neg) %>%
group_by(ID) %>%
nest()
df.NEG
## # A tibble: 1 x 2
## ID data
## <chr> <list>
## 1 ID_1 <tibble [5 × 1]>
dates.pos <- ymd(c('2018-02-07', '2018-02-12', '2018-02-13',
'2018-02-20', '2018-02-21', '2018-03-18'))
df.POS <- tibble(ID = paste0('ID_', rep(1, 6)),
DATE = dates.pos)
df.POS
## # A tibble: 6 x 2
## ID DATE
## <chr> <date>
## 1 ID_1 2018-02-07
## 2 ID_1 2018-02-12
## 3 ID_1 2018-02-13
## 4 ID_1 2018-02-20
## 5 ID_1 2018-02-21
## 6 ID_1 2018-03-18
I would like to find out for which of the positive tests there was also a negative test up to 2 days after the positive test result. I've tried using the map2() function of the purrr package
df.TOTAL <- df.POS %>%
left_join(df.NEG, by = 'ID') %>%
mutate(TIME = interval(DATE, DATE + days(2)),
RESULT = map2(data, "DATE", TIME, ~ .x %within% .y))
Unfortunaltely, my code doesn't work. The RESULT variable should be logical and return TRUE in case of a negative test result up to 2 days after the positive test. Instead it is a list and returns NULL.
df.TOTAL
## # A tibble: 6 x 5
## ID DATE data TIME RESULT
## <chr> <date> <list> <S4: Interval> <list>
## 1 ID_1 2018-02-07 <tibble [5 × 1]> 2018-02-07 UTC--2018-02-09 UTC <NULL>
## 2 ID_1 2018-02-12 <tibble [5 × 1]> 2018-02-12 UTC--2018-02-14 UTC <NULL>
## 3 ID_1 2018-02-13 <tibble [5 × 1]> 2018-02-13 UTC--2018-02-15 UTC <NULL>
## 4 ID_1 2018-02-20 <tibble [5 × 1]> 2018-02-20 UTC--2018-02-22 UTC <NULL>
## 5 ID_1 2018-02-21 <tibble [5 × 1]> 2018-02-21 UTC--2018-02-23 UTC <NULL>
## 6 ID_1 2018-03-18 <tibble [5 × 1]> 2018-03-18 UTC--2018-03-20 UTC <NULL>
Can anyone help?
I would appreciate some help. Thanks very much in advance!
r list date tidyr purrr
add a comment |
I have two data frames (tibbles) with 2 variables each:
- df.POS: ID (ID variable); DATE (Date of positive lab test)
- df.NEG: ID (ID variable); data (Date of negative lab tests (more than 1 test).
Please note that data is a list variable, created with the nest() function of the tidyr package.
library(tidyverse)
library(lubridate)
# negative tests
dates.neg <- ymd(c('2018-02-01', '2018-02-06', '2018-02-10',
'2018-02-21', '2018-04-05'))
df.NEG <- tibble(ID = paste0('ID_', rep(1, 5)),
DATE = dates.neg) %>%
group_by(ID) %>%
nest()
df.NEG
## # A tibble: 1 x 2
## ID data
## <chr> <list>
## 1 ID_1 <tibble [5 × 1]>
dates.pos <- ymd(c('2018-02-07', '2018-02-12', '2018-02-13',
'2018-02-20', '2018-02-21', '2018-03-18'))
df.POS <- tibble(ID = paste0('ID_', rep(1, 6)),
DATE = dates.pos)
df.POS
## # A tibble: 6 x 2
## ID DATE
## <chr> <date>
## 1 ID_1 2018-02-07
## 2 ID_1 2018-02-12
## 3 ID_1 2018-02-13
## 4 ID_1 2018-02-20
## 5 ID_1 2018-02-21
## 6 ID_1 2018-03-18
I would like to find out for which of the positive tests there was also a negative test up to 2 days after the positive test result. I've tried using the map2() function of the purrr package
df.TOTAL <- df.POS %>%
left_join(df.NEG, by = 'ID') %>%
mutate(TIME = interval(DATE, DATE + days(2)),
RESULT = map2(data, "DATE", TIME, ~ .x %within% .y))
Unfortunaltely, my code doesn't work. The RESULT variable should be logical and return TRUE in case of a negative test result up to 2 days after the positive test. Instead it is a list and returns NULL.
df.TOTAL
## # A tibble: 6 x 5
## ID DATE data TIME RESULT
## <chr> <date> <list> <S4: Interval> <list>
## 1 ID_1 2018-02-07 <tibble [5 × 1]> 2018-02-07 UTC--2018-02-09 UTC <NULL>
## 2 ID_1 2018-02-12 <tibble [5 × 1]> 2018-02-12 UTC--2018-02-14 UTC <NULL>
## 3 ID_1 2018-02-13 <tibble [5 × 1]> 2018-02-13 UTC--2018-02-15 UTC <NULL>
## 4 ID_1 2018-02-20 <tibble [5 × 1]> 2018-02-20 UTC--2018-02-22 UTC <NULL>
## 5 ID_1 2018-02-21 <tibble [5 × 1]> 2018-02-21 UTC--2018-02-23 UTC <NULL>
## 6 ID_1 2018-03-18 <tibble [5 × 1]> 2018-03-18 UTC--2018-03-20 UTC <NULL>
Can anyone help?
I would appreciate some help. Thanks very much in advance!
r list date tidyr purrr
add a comment |
I have two data frames (tibbles) with 2 variables each:
- df.POS: ID (ID variable); DATE (Date of positive lab test)
- df.NEG: ID (ID variable); data (Date of negative lab tests (more than 1 test).
Please note that data is a list variable, created with the nest() function of the tidyr package.
library(tidyverse)
library(lubridate)
# negative tests
dates.neg <- ymd(c('2018-02-01', '2018-02-06', '2018-02-10',
'2018-02-21', '2018-04-05'))
df.NEG <- tibble(ID = paste0('ID_', rep(1, 5)),
DATE = dates.neg) %>%
group_by(ID) %>%
nest()
df.NEG
## # A tibble: 1 x 2
## ID data
## <chr> <list>
## 1 ID_1 <tibble [5 × 1]>
dates.pos <- ymd(c('2018-02-07', '2018-02-12', '2018-02-13',
'2018-02-20', '2018-02-21', '2018-03-18'))
df.POS <- tibble(ID = paste0('ID_', rep(1, 6)),
DATE = dates.pos)
df.POS
## # A tibble: 6 x 2
## ID DATE
## <chr> <date>
## 1 ID_1 2018-02-07
## 2 ID_1 2018-02-12
## 3 ID_1 2018-02-13
## 4 ID_1 2018-02-20
## 5 ID_1 2018-02-21
## 6 ID_1 2018-03-18
I would like to find out for which of the positive tests there was also a negative test up to 2 days after the positive test result. I've tried using the map2() function of the purrr package
df.TOTAL <- df.POS %>%
left_join(df.NEG, by = 'ID') %>%
mutate(TIME = interval(DATE, DATE + days(2)),
RESULT = map2(data, "DATE", TIME, ~ .x %within% .y))
Unfortunaltely, my code doesn't work. The RESULT variable should be logical and return TRUE in case of a negative test result up to 2 days after the positive test. Instead it is a list and returns NULL.
df.TOTAL
## # A tibble: 6 x 5
## ID DATE data TIME RESULT
## <chr> <date> <list> <S4: Interval> <list>
## 1 ID_1 2018-02-07 <tibble [5 × 1]> 2018-02-07 UTC--2018-02-09 UTC <NULL>
## 2 ID_1 2018-02-12 <tibble [5 × 1]> 2018-02-12 UTC--2018-02-14 UTC <NULL>
## 3 ID_1 2018-02-13 <tibble [5 × 1]> 2018-02-13 UTC--2018-02-15 UTC <NULL>
## 4 ID_1 2018-02-20 <tibble [5 × 1]> 2018-02-20 UTC--2018-02-22 UTC <NULL>
## 5 ID_1 2018-02-21 <tibble [5 × 1]> 2018-02-21 UTC--2018-02-23 UTC <NULL>
## 6 ID_1 2018-03-18 <tibble [5 × 1]> 2018-03-18 UTC--2018-03-20 UTC <NULL>
Can anyone help?
I would appreciate some help. Thanks very much in advance!
r list date tidyr purrr
I have two data frames (tibbles) with 2 variables each:
- df.POS: ID (ID variable); DATE (Date of positive lab test)
- df.NEG: ID (ID variable); data (Date of negative lab tests (more than 1 test).
Please note that data is a list variable, created with the nest() function of the tidyr package.
library(tidyverse)
library(lubridate)
# negative tests
dates.neg <- ymd(c('2018-02-01', '2018-02-06', '2018-02-10',
'2018-02-21', '2018-04-05'))
df.NEG <- tibble(ID = paste0('ID_', rep(1, 5)),
DATE = dates.neg) %>%
group_by(ID) %>%
nest()
df.NEG
## # A tibble: 1 x 2
## ID data
## <chr> <list>
## 1 ID_1 <tibble [5 × 1]>
dates.pos <- ymd(c('2018-02-07', '2018-02-12', '2018-02-13',
'2018-02-20', '2018-02-21', '2018-03-18'))
df.POS <- tibble(ID = paste0('ID_', rep(1, 6)),
DATE = dates.pos)
df.POS
## # A tibble: 6 x 2
## ID DATE
## <chr> <date>
## 1 ID_1 2018-02-07
## 2 ID_1 2018-02-12
## 3 ID_1 2018-02-13
## 4 ID_1 2018-02-20
## 5 ID_1 2018-02-21
## 6 ID_1 2018-03-18
I would like to find out for which of the positive tests there was also a negative test up to 2 days after the positive test result. I've tried using the map2() function of the purrr package
df.TOTAL <- df.POS %>%
left_join(df.NEG, by = 'ID') %>%
mutate(TIME = interval(DATE, DATE + days(2)),
RESULT = map2(data, "DATE", TIME, ~ .x %within% .y))
Unfortunaltely, my code doesn't work. The RESULT variable should be logical and return TRUE in case of a negative test result up to 2 days after the positive test. Instead it is a list and returns NULL.
df.TOTAL
## # A tibble: 6 x 5
## ID DATE data TIME RESULT
## <chr> <date> <list> <S4: Interval> <list>
## 1 ID_1 2018-02-07 <tibble [5 × 1]> 2018-02-07 UTC--2018-02-09 UTC <NULL>
## 2 ID_1 2018-02-12 <tibble [5 × 1]> 2018-02-12 UTC--2018-02-14 UTC <NULL>
## 3 ID_1 2018-02-13 <tibble [5 × 1]> 2018-02-13 UTC--2018-02-15 UTC <NULL>
## 4 ID_1 2018-02-20 <tibble [5 × 1]> 2018-02-20 UTC--2018-02-22 UTC <NULL>
## 5 ID_1 2018-02-21 <tibble [5 × 1]> 2018-02-21 UTC--2018-02-23 UTC <NULL>
## 6 ID_1 2018-03-18 <tibble [5 × 1]> 2018-03-18 UTC--2018-03-20 UTC <NULL>
Can anyone help?
I would appreciate some help. Thanks very much in advance!
r list date tidyr purrr
r list date tidyr purrr
asked Nov 24 '18 at 17:18
Norbert KöhlerNorbert Köhler
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1 Answer
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First, note that you can test whether any element from a vector of "negative" dates falls within the "positive" interval like so:
any(dates.neg %within% interval(dates.pos[1], dates.pos[1] + days(2)))
# [1] FALSE
This suggests the following approach using map2
-- or more usefully, map2_lgl
:
df.TOTAL <- df.POS %>%
left_join(df.NEG, by = 'ID') %>%
mutate(TIME = interval(DATE, DATE + days(2)),
RESULT = map2_lgl(data, TIME, ~any(.x$DATE %within% .y)))
# # A tibble: 6 x 5
# ID DATE data TIME RESULT
# <chr> <date> <list> <S4: Interval> <lgl>
# 1 ID_1 2018-02-07 <tibble [5 x 1]> 2018-02-07 UTC--2018-02-09 UTC FALSE
# 2 ID_1 2018-02-12 <tibble [5 x 1]> 2018-02-12 UTC--2018-02-14 UTC FALSE
# 3 ID_1 2018-02-13 <tibble [5 x 1]> 2018-02-13 UTC--2018-02-15 UTC FALSE
# 4 ID_1 2018-02-20 <tibble [5 x 1]> 2018-02-20 UTC--2018-02-22 UTC TRUE
# 5 ID_1 2018-02-21 <tibble [5 x 1]> 2018-02-21 UTC--2018-02-23 UTC TRUE
# 6 ID_1 2018-03-18 <tibble [5 x 1]> 2018-03-18 UTC--2018-03-20 UTC FALSE
Thanks to @ubutun for improving the answer.
Wouldn't bemap2_lgl(data, TIME, ~ any(.x$DATE %within% y))
more self-explanatory? Anyway - great answer, thank you for the valuable information.
– utubun
Nov 24 '18 at 19:32
@utubun: Ah, that's right -- way more straightforward. I'll edit to reflect your suggestion.
– Weihuang Wong
Nov 24 '18 at 19:37
Thanks a lot. That's great! :-)
– Norbert Köhler
Nov 24 '18 at 19:54
@NorbertKöhler: Welcome to SO, and happy to help. If this answer resolved your question, please mark it as accepted.
– Weihuang Wong
Nov 24 '18 at 20:02
add a comment |
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1 Answer
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1 Answer
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votes
First, note that you can test whether any element from a vector of "negative" dates falls within the "positive" interval like so:
any(dates.neg %within% interval(dates.pos[1], dates.pos[1] + days(2)))
# [1] FALSE
This suggests the following approach using map2
-- or more usefully, map2_lgl
:
df.TOTAL <- df.POS %>%
left_join(df.NEG, by = 'ID') %>%
mutate(TIME = interval(DATE, DATE + days(2)),
RESULT = map2_lgl(data, TIME, ~any(.x$DATE %within% .y)))
# # A tibble: 6 x 5
# ID DATE data TIME RESULT
# <chr> <date> <list> <S4: Interval> <lgl>
# 1 ID_1 2018-02-07 <tibble [5 x 1]> 2018-02-07 UTC--2018-02-09 UTC FALSE
# 2 ID_1 2018-02-12 <tibble [5 x 1]> 2018-02-12 UTC--2018-02-14 UTC FALSE
# 3 ID_1 2018-02-13 <tibble [5 x 1]> 2018-02-13 UTC--2018-02-15 UTC FALSE
# 4 ID_1 2018-02-20 <tibble [5 x 1]> 2018-02-20 UTC--2018-02-22 UTC TRUE
# 5 ID_1 2018-02-21 <tibble [5 x 1]> 2018-02-21 UTC--2018-02-23 UTC TRUE
# 6 ID_1 2018-03-18 <tibble [5 x 1]> 2018-03-18 UTC--2018-03-20 UTC FALSE
Thanks to @ubutun for improving the answer.
Wouldn't bemap2_lgl(data, TIME, ~ any(.x$DATE %within% y))
more self-explanatory? Anyway - great answer, thank you for the valuable information.
– utubun
Nov 24 '18 at 19:32
@utubun: Ah, that's right -- way more straightforward. I'll edit to reflect your suggestion.
– Weihuang Wong
Nov 24 '18 at 19:37
Thanks a lot. That's great! :-)
– Norbert Köhler
Nov 24 '18 at 19:54
@NorbertKöhler: Welcome to SO, and happy to help. If this answer resolved your question, please mark it as accepted.
– Weihuang Wong
Nov 24 '18 at 20:02
add a comment |
First, note that you can test whether any element from a vector of "negative" dates falls within the "positive" interval like so:
any(dates.neg %within% interval(dates.pos[1], dates.pos[1] + days(2)))
# [1] FALSE
This suggests the following approach using map2
-- or more usefully, map2_lgl
:
df.TOTAL <- df.POS %>%
left_join(df.NEG, by = 'ID') %>%
mutate(TIME = interval(DATE, DATE + days(2)),
RESULT = map2_lgl(data, TIME, ~any(.x$DATE %within% .y)))
# # A tibble: 6 x 5
# ID DATE data TIME RESULT
# <chr> <date> <list> <S4: Interval> <lgl>
# 1 ID_1 2018-02-07 <tibble [5 x 1]> 2018-02-07 UTC--2018-02-09 UTC FALSE
# 2 ID_1 2018-02-12 <tibble [5 x 1]> 2018-02-12 UTC--2018-02-14 UTC FALSE
# 3 ID_1 2018-02-13 <tibble [5 x 1]> 2018-02-13 UTC--2018-02-15 UTC FALSE
# 4 ID_1 2018-02-20 <tibble [5 x 1]> 2018-02-20 UTC--2018-02-22 UTC TRUE
# 5 ID_1 2018-02-21 <tibble [5 x 1]> 2018-02-21 UTC--2018-02-23 UTC TRUE
# 6 ID_1 2018-03-18 <tibble [5 x 1]> 2018-03-18 UTC--2018-03-20 UTC FALSE
Thanks to @ubutun for improving the answer.
Wouldn't bemap2_lgl(data, TIME, ~ any(.x$DATE %within% y))
more self-explanatory? Anyway - great answer, thank you for the valuable information.
– utubun
Nov 24 '18 at 19:32
@utubun: Ah, that's right -- way more straightforward. I'll edit to reflect your suggestion.
– Weihuang Wong
Nov 24 '18 at 19:37
Thanks a lot. That's great! :-)
– Norbert Köhler
Nov 24 '18 at 19:54
@NorbertKöhler: Welcome to SO, and happy to help. If this answer resolved your question, please mark it as accepted.
– Weihuang Wong
Nov 24 '18 at 20:02
add a comment |
First, note that you can test whether any element from a vector of "negative" dates falls within the "positive" interval like so:
any(dates.neg %within% interval(dates.pos[1], dates.pos[1] + days(2)))
# [1] FALSE
This suggests the following approach using map2
-- or more usefully, map2_lgl
:
df.TOTAL <- df.POS %>%
left_join(df.NEG, by = 'ID') %>%
mutate(TIME = interval(DATE, DATE + days(2)),
RESULT = map2_lgl(data, TIME, ~any(.x$DATE %within% .y)))
# # A tibble: 6 x 5
# ID DATE data TIME RESULT
# <chr> <date> <list> <S4: Interval> <lgl>
# 1 ID_1 2018-02-07 <tibble [5 x 1]> 2018-02-07 UTC--2018-02-09 UTC FALSE
# 2 ID_1 2018-02-12 <tibble [5 x 1]> 2018-02-12 UTC--2018-02-14 UTC FALSE
# 3 ID_1 2018-02-13 <tibble [5 x 1]> 2018-02-13 UTC--2018-02-15 UTC FALSE
# 4 ID_1 2018-02-20 <tibble [5 x 1]> 2018-02-20 UTC--2018-02-22 UTC TRUE
# 5 ID_1 2018-02-21 <tibble [5 x 1]> 2018-02-21 UTC--2018-02-23 UTC TRUE
# 6 ID_1 2018-03-18 <tibble [5 x 1]> 2018-03-18 UTC--2018-03-20 UTC FALSE
Thanks to @ubutun for improving the answer.
First, note that you can test whether any element from a vector of "negative" dates falls within the "positive" interval like so:
any(dates.neg %within% interval(dates.pos[1], dates.pos[1] + days(2)))
# [1] FALSE
This suggests the following approach using map2
-- or more usefully, map2_lgl
:
df.TOTAL <- df.POS %>%
left_join(df.NEG, by = 'ID') %>%
mutate(TIME = interval(DATE, DATE + days(2)),
RESULT = map2_lgl(data, TIME, ~any(.x$DATE %within% .y)))
# # A tibble: 6 x 5
# ID DATE data TIME RESULT
# <chr> <date> <list> <S4: Interval> <lgl>
# 1 ID_1 2018-02-07 <tibble [5 x 1]> 2018-02-07 UTC--2018-02-09 UTC FALSE
# 2 ID_1 2018-02-12 <tibble [5 x 1]> 2018-02-12 UTC--2018-02-14 UTC FALSE
# 3 ID_1 2018-02-13 <tibble [5 x 1]> 2018-02-13 UTC--2018-02-15 UTC FALSE
# 4 ID_1 2018-02-20 <tibble [5 x 1]> 2018-02-20 UTC--2018-02-22 UTC TRUE
# 5 ID_1 2018-02-21 <tibble [5 x 1]> 2018-02-21 UTC--2018-02-23 UTC TRUE
# 6 ID_1 2018-03-18 <tibble [5 x 1]> 2018-03-18 UTC--2018-03-20 UTC FALSE
Thanks to @ubutun for improving the answer.
edited Nov 24 '18 at 19:39
answered Nov 24 '18 at 19:15
Weihuang WongWeihuang Wong
9,7101835
9,7101835
Wouldn't bemap2_lgl(data, TIME, ~ any(.x$DATE %within% y))
more self-explanatory? Anyway - great answer, thank you for the valuable information.
– utubun
Nov 24 '18 at 19:32
@utubun: Ah, that's right -- way more straightforward. I'll edit to reflect your suggestion.
– Weihuang Wong
Nov 24 '18 at 19:37
Thanks a lot. That's great! :-)
– Norbert Köhler
Nov 24 '18 at 19:54
@NorbertKöhler: Welcome to SO, and happy to help. If this answer resolved your question, please mark it as accepted.
– Weihuang Wong
Nov 24 '18 at 20:02
add a comment |
Wouldn't bemap2_lgl(data, TIME, ~ any(.x$DATE %within% y))
more self-explanatory? Anyway - great answer, thank you for the valuable information.
– utubun
Nov 24 '18 at 19:32
@utubun: Ah, that's right -- way more straightforward. I'll edit to reflect your suggestion.
– Weihuang Wong
Nov 24 '18 at 19:37
Thanks a lot. That's great! :-)
– Norbert Köhler
Nov 24 '18 at 19:54
@NorbertKöhler: Welcome to SO, and happy to help. If this answer resolved your question, please mark it as accepted.
– Weihuang Wong
Nov 24 '18 at 20:02
Wouldn't be
map2_lgl(data, TIME, ~ any(.x$DATE %within% y))
more self-explanatory? Anyway - great answer, thank you for the valuable information.– utubun
Nov 24 '18 at 19:32
Wouldn't be
map2_lgl(data, TIME, ~ any(.x$DATE %within% y))
more self-explanatory? Anyway - great answer, thank you for the valuable information.– utubun
Nov 24 '18 at 19:32
@utubun: Ah, that's right -- way more straightforward. I'll edit to reflect your suggestion.
– Weihuang Wong
Nov 24 '18 at 19:37
@utubun: Ah, that's right -- way more straightforward. I'll edit to reflect your suggestion.
– Weihuang Wong
Nov 24 '18 at 19:37
Thanks a lot. That's great! :-)
– Norbert Köhler
Nov 24 '18 at 19:54
Thanks a lot. That's great! :-)
– Norbert Köhler
Nov 24 '18 at 19:54
@NorbertKöhler: Welcome to SO, and happy to help. If this answer resolved your question, please mark it as accepted.
– Weihuang Wong
Nov 24 '18 at 20:02
@NorbertKöhler: Welcome to SO, and happy to help. If this answer resolved your question, please mark it as accepted.
– Weihuang Wong
Nov 24 '18 at 20:02
add a comment |
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
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Post as a guest
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
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Post as a guest
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
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Sign up using Google
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Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown