[英]Assign value to dynamic column name in a function in R
I have a table:我有一张桌子:
num1<-runif(10)
num2<-runif(10)
df<-data.frame(num1,num2)
df
num1 num2
1 0.41170486 0.8198323
2 0.46594131 0.7300186
3 0.09005351 0.9667960
4 0.33373968 0.6827277
5 0.48305910 0.2254627
6 0.23514394 0.3348283
7 0.79450479 0.3101271
8 0.30476853 0.6706657
9 0.53643541 0.9761387
10 0.66568823 0.8444414
And I want to create a percentile column for each of these 2 columns respectively so the table looks like this:我想分别为这 2 列中的每一列创建一个百分位列,因此表格如下所示:
df
num1 num2. num1_percentile num2_percentile
1 0.41170486 0.8198323 20% 40%
2 0.46594131 0.7300186 xxx xxx
3 0.09005351 0.9667960 xxx xxx
4 0.33373968 0.6827277 xxx xxx
5 0.48305910 0.2254627 xxx xxx
6 0.23514394 0.3348283 xxx xxx
7 0.79450479 0.3101271 xxx xxx
8 0.30476853 0.6706657 xxx xxx
9 0.53643541 0.9761387 xxx xxx
10 0.66568823 0.8444414 xxx xxx
So the percentile column name will be based on the if it's num1
or num2
.因此,百分位列名称将基于它是num1
还是num2
。 I have this function where I am trying to assign percentile values to dynamic column names but seems like assign()
did not store values
in num1_percentile
.我有这个 function 我试图将百分位值分配给动态列名,但似乎assign()
没有将values
存储在num1_percentile
中。
create_percentile<-function(num){
breaks=c( quantile(df[[num]], probs =seq(0, 1, 0.2)))
values<- cut(df[[num]],
breaks=breaks,labels=c( '20%','40%', '60%','80%','100%'))
values[is.na(values)]<-'20%'
print(values)
assign(paste0(df,'_percentile'),values)
df<-cbind(df,paste0(num,'_percentile'))
return(df)
}
create_percentile('num1')
How do I make it work so it can assign values to dynamic column names?如何使它工作,以便它可以为动态列名分配值?
The reason behind assign
not storing the value is that you are storing it in assign(paste0(df,'_percentile'),values)
where df is a data frame. assign
不存储值的原因是您将其存储在assign(paste0(df,'_percentile'),values)
中,其中 df 是数据框。 So while assigning it considers the first element of df to paste it with _percentile .因此,在分配它时会考虑将 df 的第一个元素与_percentile一起粘贴。 A small correction is instead of df
you can use num
.您可以使用num
来代替df
进行小的更正。
Also this function cbind(df,paste0(num,'_percentile'))
won't work as expected because it will store a character value returned from paste
command.此 function cbind(df,paste0(num,'_percentile'))
也不会按预期工作,因为它将存储从paste
命令返回的字符值。
You may use the below code to assign the values to dynamic column names您可以使用以下代码将值分配给动态列名
num1 <- runif(10)
num2 <- runif(10)
df <- data.frame(num1, num2)
df
num1 num2
1 0.78584180 0.61999953
2 0.33916132 0.91999116
3 0.84080205 0.19279828
4 0.03563479 0.04253741
5 0.16633144 0.23509481
6 0.15459396 0.84458053
7 0.99544676 0.08727064
8 0.25060605 0.51908750
9 0.35631173 0.48853416
10 0.50649402 0.64245468
create_percentile <- function(num) {
breaks <- quantile(df[[num]], probs = seq(0, 1, 0.2))
values <- cut(df[[num]], breaks = breaks, labels = c('20%', '40%', '60%', '80%', '100%'))
values[is.na(values)] <- '20%'
df <- cbind(df, values)
names(df)[ncol(df)] <- paste0(num, '_percentile')
return(df)
}
df <- create_percentile('num1')
df <- create_percentile('num2')
Output: Output:
num1 num2 num1_percentile num2_percentile
1 0.78584180 0.61999953 80% 80%
2 0.33916132 0.91999116 60% 100%
3 0.84080205 0.19279828 100% 40%
4 0.03563479 0.04253741 20% 20%
5 0.16633144 0.23509481 40% 40%
6 0.15459396 0.84458053 20% 100%
7 0.99544676 0.08727064 100% 20%
8 0.25060605 0.51908750 40% 60%
9 0.35631173 0.48853416 60% 60%
10 0.50649402 0.64245468 80% 80%
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