How to get the name of a Spark Column as String?
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I want to write a method to round a numeric column without doing something like:
df
.select(round($"x",2).as("x"))
Therefore I need to have a reusable column-expression like:
def roundKeepName(c:Column,scale:Int) = round(c,scale).as(c.name)
Unfortunately c.name does not exist, therefore the above code does not compile. I've found a solution for ColumName:
def roundKeepName(c:ColumnName,scale:Int) = round(c,scale).as(c.string.name)
But how can I do that with Column (which is generated if I use col("x") instead of $"x")
scala apache-spark
edited Nov 26 '18 at 16:01
Bugs
4,15992637
asked Nov 26 '18 at 11:00
Raphael RothRaphael Roth
12.8k54279
add a comment |
0
I want to write a method to round a numeric column without doing something like:
df
.select(round($"x",2).as("x"))
Therefore I need to have a reusable column-expression like:
def roundKeepName(c:Column,scale:Int) = round(c,scale).as(c.name)
Unfortunately c.name does not exist, therefore the above code does not compile. I've found a solution for ColumName:
def roundKeepName(c:ColumnName,scale:Int) = round(c,scale).as(c.string.name)
But how can I do that with Column (which is generated if I use col("x") instead of $"x")
scala apache-spark
edited Nov 26 '18 at 16:01
Bugs
4,15992637
asked Nov 26 '18 at 11:00
Raphael RothRaphael Roth
12.8k54279
c.expr.toString?
– user10465355
Nov 26 '18 at 11:08
add a comment |
0
0
0
0
I want to write a method to round a numeric column without doing something like:
df
.select(round($"x",2).as("x"))
Therefore I need to have a reusable column-expression like:
def roundKeepName(c:Column,scale:Int) = round(c,scale).as(c.name)
Unfortunately c.name does not exist, therefore the above code does not compile. I've found a solution for ColumName:
def roundKeepName(c:ColumnName,scale:Int) = round(c,scale).as(c.string.name)
But how can I do that with Column (which is generated if I use col("x") instead of $"x")
scala apache-spark
edited Nov 26 '18 at 16:01
Bugs
4,15992637
asked Nov 26 '18 at 11:00
Raphael RothRaphael Roth
12.8k54279
I want to write a method to round a numeric column without doing something like:
df
.select(round($"x",2).as("x"))
Therefore I need to have a reusable column-expression like:
def roundKeepName(c:Column,scale:Int) = round(c,scale).as(c.name)
Unfortunately c.name does not exist, therefore the above code does not compile. I've found a solution for ColumName:
def roundKeepName(c:ColumnName,scale:Int) = round(c,scale).as(c.string.name)
But how can I do that with Column (which is generated if I use col("x") instead of $"x")
scala apache-spark
scala apache-spark
edited Nov 26 '18 at 16:01
Bugs
4,15992637
asked Nov 26 '18 at 11:00
Raphael RothRaphael Roth
12.8k54279
edited Nov 26 '18 at 16:01
Bugs
4,15992637
asked Nov 26 '18 at 11:00
Raphael RothRaphael Roth
12.8k54279
edited Nov 26 '18 at 16:01
Bugs
4,15992637
edited Nov 26 '18 at 16:01
Bugs
4,15992637
edited Nov 26 '18 at 16:01
Bugs
4,15992637
4,15992637
asked Nov 26 '18 at 11:00
Raphael RothRaphael Roth
12.8k54279
asked Nov 26 '18 at 11:00
Raphael RothRaphael Roth
12.8k54279
asked Nov 26 '18 at 11:00
Raphael RothRaphael Roth
12.8k54279
12.8k54279
c.expr.toString?
– user10465355
Nov 26 '18 at 11:08
add a comment |
c.expr.toString?
– user10465355
Nov 26 '18 at 11:08
c.expr.toString?– user10465355
Nov 26 '18 at 11:08
c.expr.toString?– user10465355
Nov 26 '18 at 11:08
add a comment |
2 Answers
2
active
oldest
votes
1
Not sure if the question has really been answered. Your function could be implemented like this (toString returns the name of the column):
def roundKeepname(c:Column,scale:Int) = round(c,scale).as(c.toString)
In case you don't like relying on toString, here is a more robust version. You can rely on the underlying expression, cast it to a NamedExpression and take its name.
import org.apache.spark.sql.catalyst.expressions.NamedExpression
def roundKeepname(c:Column,scale:Int) =
c.expr.asInstanceOf[NamedExpression].name
And it works:
scala> spark.range(2).select(roundKeepname('id, 2)).show
+---+
| id|
+---+
| 0|
| 1|
+---+
answered Nov 26 '18 at 14:34
OliOli
2,3141419
that works, altough I don't like to rely on.toString
– Raphael Roth
Nov 26 '18 at 15:41
Could you tell us why you need to define your function from the column object? I usually do things like that from the column name withround(col(name), 2).as(name). Would it work for you?
– Oli
Nov 26 '18 at 18:02
I edited my answer with another more robust version
– Oli
Nov 26 '18 at 18:28
thats a good idea, you meandef roundKeepName(columnName:String,scale:Int) = round(col(columnName),scale).as(columnName)?
– Raphael Roth
Nov 27 '18 at 8:24
Yes exactly. This is something I use quite often.
– Oli
Nov 27 '18 at 9:24
add a comment |
-1
Update:
With the solution way given by BlueSheepToken, here is how you can do it dynamically assuming you have all "double" columns.
scala> val df = Seq((1.22,4.34,8.93),(3.44,12.66,17.44),(5.66,9.35,6.54)).toDF("x","y","z")
df: org.apache.spark.sql.DataFrame = [x: double, y: double ... 1 more field]
scala> df.show
+----+-----+-----+
| x| y| z|
+----+-----+-----+
|1.22| 4.34| 8.93|
|3.44|12.66|17.44|
|5.66| 9.35| 6.54|
+----+-----+-----+
scala> df.columns.foldLeft(df)( (acc,p) => (acc.withColumn(p+"_t",round(col(p),1)).drop(p).withColumnRenamed(p+"_t",p))).show
+---+----+----+
| x| y| z|
+---+----+----+
|1.2| 4.3| 8.9|
|3.4|12.7|17.4|
|5.7| 9.4| 6.5|
+---+----+----+
scala>
answered Nov 26 '18 at 12:49
stack0114106stack0114106
4,9832423
Is it me or you forgot to drop the x column before renaming? :)
– BlueSheepToken
Nov 26 '18 at 12:57
if you drop x first, then you will get exception - cannot resolve 'x' .. df.drop("x").withColumn("x2",round($"x",1)).show(false) doesn't work.
– stack0114106
Nov 26 '18 at 13:00
Ok my bad then, I thoughtdf.withColumn("x2",round($"x",1)).drop("x").withColumnRenamed("x2","x").showwas working
– BlueSheepToken
Nov 26 '18 at 13:23
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
Not sure if the question has really been answered. Your function could be implemented like this (toString returns the name of the column):
def roundKeepname(c:Column,scale:Int) = round(c,scale).as(c.toString)
In case you don't like relying on toString, here is a more robust version. You can rely on the underlying expression, cast it to a NamedExpression and take its name.
import org.apache.spark.sql.catalyst.expressions.NamedExpression
def roundKeepname(c:Column,scale:Int) =
c.expr.asInstanceOf[NamedExpression].name
And it works:
scala> spark.range(2).select(roundKeepname('id, 2)).show
+---+
| id|
+---+
| 0|
| 1|
+---+
answered Nov 26 '18 at 14:34
OliOli
2,3141419
that works, altough I don't like to rely on.toString
– Raphael Roth
Nov 26 '18 at 15:41
Could you tell us why you need to define your function from the column object? I usually do things like that from the column name withround(col(name), 2).as(name). Would it work for you?
– Oli
Nov 26 '18 at 18:02
I edited my answer with another more robust version
– Oli
Nov 26 '18 at 18:28
thats a good idea, you meandef roundKeepName(columnName:String,scale:Int) = round(col(columnName),scale).as(columnName)?
– Raphael Roth
Nov 27 '18 at 8:24
Yes exactly. This is something I use quite often.
– Oli
Nov 27 '18 at 9:24
add a comment |
1
Not sure if the question has really been answered. Your function could be implemented like this (toString returns the name of the column):
def roundKeepname(c:Column,scale:Int) = round(c,scale).as(c.toString)
In case you don't like relying on toString, here is a more robust version. You can rely on the underlying expression, cast it to a NamedExpression and take its name.
import org.apache.spark.sql.catalyst.expressions.NamedExpression
def roundKeepname(c:Column,scale:Int) =
c.expr.asInstanceOf[NamedExpression].name
And it works:
scala> spark.range(2).select(roundKeepname('id, 2)).show
+---+
| id|
+---+
| 0|
| 1|
+---+
answered Nov 26 '18 at 14:34
OliOli
2,3141419
that works, altough I don't like to rely on.toString
– Raphael Roth
Nov 26 '18 at 15:41
Could you tell us why you need to define your function from the column object? I usually do things like that from the column name withround(col(name), 2).as(name). Would it work for you?
– Oli
Nov 26 '18 at 18:02
I edited my answer with another more robust version
– Oli
Nov 26 '18 at 18:28
thats a good idea, you meandef roundKeepName(columnName:String,scale:Int) = round(col(columnName),scale).as(columnName)?
– Raphael Roth
Nov 27 '18 at 8:24
Yes exactly. This is something I use quite often.
– Oli
Nov 27 '18 at 9:24
add a comment |
1
1
1
Not sure if the question has really been answered. Your function could be implemented like this (toString returns the name of the column):
def roundKeepname(c:Column,scale:Int) = round(c,scale).as(c.toString)
In case you don't like relying on toString, here is a more robust version. You can rely on the underlying expression, cast it to a NamedExpression and take its name.
import org.apache.spark.sql.catalyst.expressions.NamedExpression
def roundKeepname(c:Column,scale:Int) =
c.expr.asInstanceOf[NamedExpression].name
And it works:
scala> spark.range(2).select(roundKeepname('id, 2)).show
+---+
| id|
+---+
| 0|
| 1|
+---+
answered Nov 26 '18 at 14:34
OliOli
2,3141419
Not sure if the question has really been answered. Your function could be implemented like this (toString returns the name of the column):
def roundKeepname(c:Column,scale:Int) = round(c,scale).as(c.toString)
In case you don't like relying on toString, here is a more robust version. You can rely on the underlying expression, cast it to a NamedExpression and take its name.
import org.apache.spark.sql.catalyst.expressions.NamedExpression
def roundKeepname(c:Column,scale:Int) =
c.expr.asInstanceOf[NamedExpression].name
And it works:
scala> spark.range(2).select(roundKeepname('id, 2)).show
+---+
| id|
+---+
| 0|
| 1|
+---+
answered Nov 26 '18 at 14:34
OliOli
2,3141419
edited Nov 26 '18 at 18:28
answered Nov 26 '18 at 14:34
OliOli
2,3141419
answered Nov 26 '18 at 14:34
OliOli
2,3141419
answered Nov 26 '18 at 14:34
OliOli
2,3141419
2,3141419
that works, altough I don't like to rely on.toString
– Raphael Roth
Nov 26 '18 at 15:41
Could you tell us why you need to define your function from the column object? I usually do things like that from the column name withround(col(name), 2).as(name). Would it work for you?
– Oli
Nov 26 '18 at 18:02
I edited my answer with another more robust version
– Oli
Nov 26 '18 at 18:28
thats a good idea, you meandef roundKeepName(columnName:String,scale:Int) = round(col(columnName),scale).as(columnName)?
– Raphael Roth
Nov 27 '18 at 8:24
Yes exactly. This is something I use quite often.
– Oli
Nov 27 '18 at 9:24
add a comment |
that works, altough I don't like to rely on.toString
– Raphael Roth
Nov 26 '18 at 15:41
Could you tell us why you need to define your function from the column object? I usually do things like that from the column name withround(col(name), 2).as(name). Would it work for you?
– Oli
Nov 26 '18 at 18:02
I edited my answer with another more robust version
– Oli
Nov 26 '18 at 18:28
thats a good idea, you meandef roundKeepName(columnName:String,scale:Int) = round(col(columnName),scale).as(columnName)?
– Raphael Roth
Nov 27 '18 at 8:24
Yes exactly. This is something I use quite often.
– Oli
Nov 27 '18 at 9:24
that works, altough I don't like to rely on
.toString– Raphael Roth
Nov 26 '18 at 15:41
that works, altough I don't like to rely on
.toString– Raphael Roth
Nov 26 '18 at 15:41
Could you tell us why you need to define your function from the column object? I usually do things like that from the column name with
round(col(name), 2).as(name). Would it work for you?– Oli
Nov 26 '18 at 18:02
Could you tell us why you need to define your function from the column object? I usually do things like that from the column name with
round(col(name), 2).as(name). Would it work for you?– Oli
Nov 26 '18 at 18:02
I edited my answer with another more robust version
– Oli
Nov 26 '18 at 18:28
I edited my answer with another more robust version
– Oli
Nov 26 '18 at 18:28
thats a good idea, you mean
def roundKeepName(columnName:String,scale:Int) = round(col(columnName),scale).as(columnName)?– Raphael Roth
Nov 27 '18 at 8:24
thats a good idea, you mean
def roundKeepName(columnName:String,scale:Int) = round(col(columnName),scale).as(columnName)?– Raphael Roth
Nov 27 '18 at 8:24
Yes exactly. This is something I use quite often.
– Oli
Nov 27 '18 at 9:24
Yes exactly. This is something I use quite often.
– Oli
Nov 27 '18 at 9:24
add a comment |
-1
Update:
With the solution way given by BlueSheepToken, here is how you can do it dynamically assuming you have all "double" columns.
scala> val df = Seq((1.22,4.34,8.93),(3.44,12.66,17.44),(5.66,9.35,6.54)).toDF("x","y","z")
df: org.apache.spark.sql.DataFrame = [x: double, y: double ... 1 more field]
scala> df.show
+----+-----+-----+
| x| y| z|
+----+-----+-----+
|1.22| 4.34| 8.93|
|3.44|12.66|17.44|
|5.66| 9.35| 6.54|
+----+-----+-----+
scala> df.columns.foldLeft(df)( (acc,p) => (acc.withColumn(p+"_t",round(col(p),1)).drop(p).withColumnRenamed(p+"_t",p))).show
+---+----+----+
| x| y| z|
+---+----+----+
|1.2| 4.3| 8.9|
|3.4|12.7|17.4|
|5.7| 9.4| 6.5|
+---+----+----+
scala>
answered Nov 26 '18 at 12:49
stack0114106stack0114106
4,9832423
Is it me or you forgot to drop the x column before renaming? :)
– BlueSheepToken
Nov 26 '18 at 12:57
if you drop x first, then you will get exception - cannot resolve 'x' .. df.drop("x").withColumn("x2",round($"x",1)).show(false) doesn't work.
– stack0114106
Nov 26 '18 at 13:00
Ok my bad then, I thoughtdf.withColumn("x2",round($"x",1)).drop("x").withColumnRenamed("x2","x").showwas working
– BlueSheepToken
Nov 26 '18 at 13:23
add a comment |
-1
Update:
With the solution way given by BlueSheepToken, here is how you can do it dynamically assuming you have all "double" columns.
scala> val df = Seq((1.22,4.34,8.93),(3.44,12.66,17.44),(5.66,9.35,6.54)).toDF("x","y","z")
df: org.apache.spark.sql.DataFrame = [x: double, y: double ... 1 more field]
scala> df.show
+----+-----+-----+
| x| y| z|
+----+-----+-----+
|1.22| 4.34| 8.93|
|3.44|12.66|17.44|
|5.66| 9.35| 6.54|
+----+-----+-----+
scala> df.columns.foldLeft(df)( (acc,p) => (acc.withColumn(p+"_t",round(col(p),1)).drop(p).withColumnRenamed(p+"_t",p))).show
+---+----+----+
| x| y| z|
+---+----+----+
|1.2| 4.3| 8.9|
|3.4|12.7|17.4|
|5.7| 9.4| 6.5|
+---+----+----+
scala>
answered Nov 26 '18 at 12:49
stack0114106stack0114106
4,9832423
Is it me or you forgot to drop the x column before renaming? :)
– BlueSheepToken
Nov 26 '18 at 12:57
if you drop x first, then you will get exception - cannot resolve 'x' .. df.drop("x").withColumn("x2",round($"x",1)).show(false) doesn't work.
– stack0114106
Nov 26 '18 at 13:00
Ok my bad then, I thoughtdf.withColumn("x2",round($"x",1)).drop("x").withColumnRenamed("x2","x").showwas working
– BlueSheepToken
Nov 26 '18 at 13:23
add a comment |
-1
-1
-1
Update:
With the solution way given by BlueSheepToken, here is how you can do it dynamically assuming you have all "double" columns.
scala> val df = Seq((1.22,4.34,8.93),(3.44,12.66,17.44),(5.66,9.35,6.54)).toDF("x","y","z")
df: org.apache.spark.sql.DataFrame = [x: double, y: double ... 1 more field]
scala> df.show
+----+-----+-----+
| x| y| z|
+----+-----+-----+
|1.22| 4.34| 8.93|
|3.44|12.66|17.44|
|5.66| 9.35| 6.54|
+----+-----+-----+
scala> df.columns.foldLeft(df)( (acc,p) => (acc.withColumn(p+"_t",round(col(p),1)).drop(p).withColumnRenamed(p+"_t",p))).show
+---+----+----+
| x| y| z|
+---+----+----+
|1.2| 4.3| 8.9|
|3.4|12.7|17.4|
|5.7| 9.4| 6.5|
+---+----+----+
scala>
answered Nov 26 '18 at 12:49
stack0114106stack0114106
4,9832423
Update:
With the solution way given by BlueSheepToken, here is how you can do it dynamically assuming you have all "double" columns.
scala> val df = Seq((1.22,4.34,8.93),(3.44,12.66,17.44),(5.66,9.35,6.54)).toDF("x","y","z")
df: org.apache.spark.sql.DataFrame = [x: double, y: double ... 1 more field]
scala> df.show
+----+-----+-----+
| x| y| z|
+----+-----+-----+
|1.22| 4.34| 8.93|
|3.44|12.66|17.44|
|5.66| 9.35| 6.54|
+----+-----+-----+
scala> df.columns.foldLeft(df)( (acc,p) => (acc.withColumn(p+"_t",round(col(p),1)).drop(p).withColumnRenamed(p+"_t",p))).show
+---+----+----+
| x| y| z|
+---+----+----+
|1.2| 4.3| 8.9|
|3.4|12.7|17.4|
|5.7| 9.4| 6.5|
+---+----+----+
scala>
answered Nov 26 '18 at 12:49
stack0114106stack0114106
4,9832423
edited Nov 26 '18 at 14:29
answered Nov 26 '18 at 12:49
stack0114106stack0114106
4,9832423
answered Nov 26 '18 at 12:49
stack0114106stack0114106
4,9832423
answered Nov 26 '18 at 12:49
stack0114106stack0114106
4,9832423
4,9832423
Is it me or you forgot to drop the x column before renaming? :)
– BlueSheepToken
Nov 26 '18 at 12:57
if you drop x first, then you will get exception - cannot resolve 'x' .. df.drop("x").withColumn("x2",round($"x",1)).show(false) doesn't work.
– stack0114106
Nov 26 '18 at 13:00
Ok my bad then, I thoughtdf.withColumn("x2",round($"x",1)).drop("x").withColumnRenamed("x2","x").showwas working
– BlueSheepToken
Nov 26 '18 at 13:23
add a comment |
Is it me or you forgot to drop the x column before renaming? :)
– BlueSheepToken
Nov 26 '18 at 12:57
if you drop x first, then you will get exception - cannot resolve 'x' .. df.drop("x").withColumn("x2",round($"x",1)).show(false) doesn't work.
– stack0114106
Nov 26 '18 at 13:00
Ok my bad then, I thoughtdf.withColumn("x2",round($"x",1)).drop("x").withColumnRenamed("x2","x").showwas working
– BlueSheepToken
Nov 26 '18 at 13:23
Is it me or you forgot to drop the x column before renaming? :)
– BlueSheepToken
Nov 26 '18 at 12:57
Is it me or you forgot to drop the x column before renaming? :)
– BlueSheepToken
Nov 26 '18 at 12:57
if you drop x first, then you will get exception - cannot resolve '
x' .. df.drop("x").withColumn("x2",round($"x",1)).show(false) doesn't work.– stack0114106
Nov 26 '18 at 13:00
if you drop x first, then you will get exception - cannot resolve '
x' .. df.drop("x").withColumn("x2",round($"x",1)).show(false) doesn't work.– stack0114106
Nov 26 '18 at 13:00
Ok my bad then, I thought
df.withColumn("x2",round($"x",1)).drop("x").withColumnRenamed("x2","x").show was working– BlueSheepToken
Nov 26 '18 at 13:23
Ok my bad then, I thought
df.withColumn("x2",round($"x",1)).drop("x").withColumnRenamed("x2","x").show was working– BlueSheepToken
Nov 26 '18 at 13:23
add a comment |
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c.expr.toString?– user10465355
Nov 26 '18 at 11:08