按名称减去数据框的列

编程问答 2022-06-12

问题描述

假设我有一个如下所示的数据框：

df <- as.data.frame(matrix(seq(1,20,1),nrow=4),byrow=TRUE)
colnames(df) <- c("X1","X2","X3","X4","X5")
rownames(df) <- as.Date(c("2020-01-02","2020-01-03","2020-01-04","2020-01-05"))

df
           X1 X2 X3 X4 X5
2020-01-02  1  2  3  4  5
2020-01-03  6  7  8  9 10
2020-01-04 11 12 13 14 15
2020-01-05 16 17 18 19 20

我想从第一列X1中减去所有列，并将其存储在同一列中。我已经尝试过了

  for(i in colnames(df)){
    df[i] <- lapply(df[i],function(x) x-df["X1"])
  }

但是它仅将其应用于第一列。如何为所有列运行它？

解决方法

尝试此base R解决方案，不要循环。只要记住列的位置：

#Data
df <- as.data.frame(matrix(seq(1,20,1),nrow=4),byrow=TRUE)
colnames(df) <- c("X1","X2","X3","X4","X5")
rownames(df) <- as.Date(c("2020-01-02","2020-01-03","2020-01-04","2020-01-05"))
#Set columns for difference
df[,2:5] <- df[,2:5]-df[,1]

输出：

           X1 X2 X3 X4 X5
2020-01-02  1  4  8 12 16
2020-01-03  2  4  8 12 16
2020-01-04  3  4  8 12 16
2020-01-05  4  4  8 12 16

或更复杂的方法是：

#Create index
#Var to substract
i1 <- which(names(df)=='X1')
#Vars to be substracted with X1
i2 <- which(names(df)!='X1')
#Compute
df[,i2]<-df[,i2]-df[,i1]

输出：

           X1 X2 X3 X4 X5
2020-01-02  1  4  8 12 16
2020-01-03  2  4  8 12 16
2020-01-04  3  4  8 12 16
2020-01-05  4  4  8 12 16
,

如果您想坚持lapply，可以这样做：

df[] <- lapply(df,`-`,df$X1)
df
#            X1 X2 X3 X4 X5
# 2020-01-02  0  4  8 12 16
# 2020-01-03  0  4  8 12 16
# 2020-01-04  0  4  8 12 16
# 2020-01-05  0  4  8 12 16
,

这是grep的一种方式：

i_col <- grep("X1",names(df))
df[] <- df - df[,i_col]
df
#           X1 X2 X3 X4 X5
#2020-01-02  0  4  8 12 16
#2020-01-03  0  4  8 12 16
#2020-01-04  0  4  8 12 16
#2020-01-05  0  4  8 12 16

另一个是grep/sweep。实际上，-是sweep的默认功能。

sweep(df,1,df[[i_col]],check.margin = FALSE)
#           X1 X2 X3 X4 X5
#2020-01-02  0  4  8 12 16
#2020-01-03  0  4  8 12 16
#2020-01-04  0  4  8 12 16
#2020-01-05  0  4  8 12 16
dataframerrsubtraction