我有一个包含多列的数据框
col1|col2|col3|colA|colB|colC|Percent
1 1 1 2 2 2 50
之前我将这些列作为子集并创建了一个向量
ColAlphabet<-c("ColA","ColB","ColC")
我想做的是将ColAlphabet乘以Percent,这样最终我就可以了
col1|col2|col3|colA|colB|colC|Percent
1 1 1 1 1 1 50
答案 0 :(得分:1)
我们可以将mutate
与across
一起使用。指定用all_of
包裹的感兴趣的列,并将这些列乘以“百分比”
library(dplyr)
df2 <- df1 %>%
mutate(across(all_of(ColAlphabet), ~ .* Percent/100))
-输出
df2
# col1 col2 col3 colA colB colC Percent
#1 1 1 1 1 1 1 50
df1 <- structure(list(col1 = 1L, col2 = 1L, col3 = 1L, colA = 2L, colB = 2L,
colC = 2L, Percent = 50L), class = "data.frame", row.names = c(NA,
-1L))
答案 1 :(得分:1)
您可以将该列作为子集,乘以Percent
并将其再次保存在ColAlphabet
中。
ColAlphabet<-c("colA","colB","colC")
df[ColAlphabet] <- df[ColAlphabet] * df$Percent/100
df
# col1 col2 col3 colA colB colC Percent
#1 1 1 1 1 1 1 50
答案 2 :(得分:0)
我们还可以使用apply()
:
#Vector
ColAlphabet<-c("colA","colB","colC")
#Code
df[,ColAlphabet] <- apply(df[,ColAlphabet],2,function(x) x*df$Percent/100)
输出:
df
col1 col2 col3 colA colB colC Percent
1 1 1 1 1 1 1 50
使用了一些数据:
#Data
df <- structure(list(col1 = 1L, col2 = 1L, col3 = 1L, colA = 2L, colB = 2L,
colC = 2L, Percent = 50L), class = "data.frame", row.names = c(NA,
-1L))
答案 3 :(得分:0)
如果您想直接相乘:
> df <- data.frame(col1 = 1, col2 = 1, col3 = 1, colA = 2, colB = 2, colC = 2, Percent = 50)
> df
col1 col2 col3 colA colB colC Percent
1 1 1 1 2 2 2 50
> df[grep('^c.*[A-Z]$', names(df))] <- df[grep('^c.*[A-Z]$', names(df))] * df$Percent/100
> df
col1 col2 col3 colA colB colC Percent
1 1 1 1 1 1 1 50
>