r - R中同一数据框中的多列相减

Question

对于以下数据集

mydata=data.frame(x1_c1=c(1:5),
                  x2_c1=c(2:6),
                  x3_c1=c(3:7),
                  x4_c1=c(4:8),
                  x1_c2=0,
                  x2_c2=0,
                  x3_c2=0,
                  x4_c2=0,
                  x1_c3=c(1:5),
                  x2_c3=c(2:6),
                  x3_c3=c(3:7),
                  x4_c3=c(4:8))

> mydata
  x1_c1 x2_c1 x3_c1 x4_c1 x1_c2 x2_c2 x3_c2 x4_c2 x1_c3 x2_c3 x3_c3 x4_c3
1     1     2     3     4     0     0     0     0     1     2     3     4
2     2     3     4     5     0     0     0     0     2     3     4     5
3     3     4     5     6     0     0     0     0     3     4     5     6
4     4     5     6     7     0     0     0     0     4     5     6     7
5     5     6     7     8     0     0     0     0     5     6     7     8

我想从以 _c1、_c2 和 _c3 结尾的变量中减去以 _c3 结尾的所有变量 code> 然后合并所有列。这是一个执行此操作的选项

mydata_update=cbind(mydata[,grep("_c1", colnames(mydata)) ]-mydata[,grep("_c3", colnames(mydata)) ],
                    mydata[,grep("_c2", colnames(mydata)) ]-mydata[,grep("_c3", colnames(mydata)) ],
                    mydata[,grep("_c3", colnames(mydata)) ]-mydata[,grep("_c3", colnames(mydata)) ])

预期结果是

> mydata_update
  x1_c1 x2_c1 x3_c1 x4_c1 x1_c2 x2_c2 x3_c2 x4_c2 x1_c3 x2_c3 x3_c3 x4_c3
1     0     0     0     0    -1    -2    -3    -4     0     0     0     0
2     0     0     0     0    -2    -3    -4    -5     0     0     0     0
3     0     0     0     0    -3    -4    -5    -6     0     0     0     0
4     0     0     0     0    -4    -5    -6    -7     0     0     0     0
5     0     0     0     0    -5    -6    -7    -8     0     0     0     0

欢迎任何其他方法。

Answer 1

匹配数据和减法部分的前缀，然后减去:

subsel <- endsWith(names(mydata), "_c3")
prefix <- sub("_.+", "", names(mydata))
mydata - mydata[subsel][match(prefix, prefix[subsel])]

#  x1_c1 x2_c1 x3_c1 x4_c1 x1_c2 x2_c2 x3_c2 x4_c2 x1_c3 x2_c3 x3_c3 x4_c3
#1     0     0     0     0    -1    -2    -3    -4     0     0     0     0
#2     0     0     0     0    -2    -3    -4    -5     0     0     0     0
#3     0     0     0     0    -3    -4    -5    -6     0     0     0     0
#4     0     0     0     0    -4    -5    -6    -7     0     0     0     0
#5     0     0     0     0    -5    -6    -7    -8     0     0     0     0

或者，如果您想生活在边缘，并且确定您的数据完整且按预期排序:

mydata - as.matrix(mydata[,endsWith(names(mydata), "_c3")])