在數據幀上的列名的基礎添加兩列

Question

實施例：我有一個數據幀

> a = data.frame(T_a_1=c(1,2,3,4,5),T_a_2=c(2,3,4,5,6),T_b_1=c(3,4,5,6,7),T_c_1=c(4,5,6,7,8),length=c(1,2,3,4,5)) 
> a  
T_a_1 T_a_2 T_b_1 T_c_1 length 
1  2  3  4  1 
2  3  4  5  2 
3  4  5  6  3 
4  5  6  7  4 
5  6  7  8  5 

我要添加（或做類似（列1 +列2）/長度上的基礎上，列一些其它操作名字。 像T_A（T_a_1和T_a_2）是兩列（第一和第二）之間共同的名字，所以我想補充它們。

Answer 1

我會用grep命令作業列名對一些模式相匹配。下面是一些例子：

> a = data.frame(T_a_1=c(1,2,3,4,5), 
+    T_a_2=c(2,3,4,5,6), 
+    T_b_1=c(3,4,5,6,7), 
+    T_c_1=c(4,5,6,7,8), 
+    length=c(1,2,3,4,5)) 
> 
> # display only columns that match T_a 
> a[,grep('T_a', colnames(a))] 
    T_a_1 T_a_2 
1  1  2 
2  2  3 
3  3  4 
4  4  5 
5  5  6 
> 
> # sum 
> sum(a[,grep('T_a', colnames(a))]) 
[1] 35 
> 
> #rowsum 
> rowSums(a[,grep('T_a', colnames(a))]) 
[1] 3 5 7 9 11 
> 
> # your example (row1 + row2)/length 
> rowSums(a[,grep('T_a', colnames(a))])/a$length 
[1] 3.000000 2.500000 2.333333 2.250000 2.200000 

更新：

從下面的評論，我明白你想總結匹配的行按常用前綴分組和長度分割列。以下代碼是用於該問題的不雅溶液：

> a = data.frame(ES51_223_1=c(1,2,3,4,5), 
+    ES51_312_1=c(2,3,4,5,6), 
+    ES52_223_2=c(3,4,5,6,7), 
+    ES52_312_2=c(4,5,6,7,8), 
+    ES53_223_3=c(1,2,3,4,5), 
+    length=c(1,2,3,4,5)) 
> 
> # get the unique prefixes 
> prefixes = unique(unlist(lapply(colnames(subset(a, select=-length)), function(x) { strsplit(x, '_')[[1]][[1]]}))) 
> 
> f = function(prefix) { 
+ return (rowSums(subset(a, select=grep(prefix, colnames(a))))/a$length) 
+ } 
> m = matrix(unlist(lapply(prefixes, f)), nrow=nrow(a)) 
> colnames(m) = prefixes 
> m 
     ES51  ES52 ES53 
[1,] 3.000000 7.000000 1 
[2,] 2.500000 4.500000 1 
[3,] 2.333333 3.666667 1 
[4,] 2.250000 3.250000 1 
[5,] 2.200000 3.000000 1 

m是包含在不同的列不同的前綴的結果矩陣。

Answer 2

這個怎麼樣？

# data 
df <- structure(list(ES51_223_1 = 1:5, ES51_312_1 = 2:6, ES52_223_2 = 3:7, 
     ES52_312_2 = 4:8, ES53_223_3 = 1:5, length = 1:5), 
     .Names = c("ES51_223_1", "ES51_312_1", "ES52_223_2", "ES52_312_2", 
     "ES53_223_3", "length"), row.names = c(NA, -5L), class = "data.frame") 

# create indices from factor levels (shortcut) 
ids <- gsub("_.*$", "", setdiff(names(df), "length")) 
ids <- factor(as.numeric(factor(ids))) 
> ids 
# [1] 1 1 2 2 3 
# Levels: 1 2 3 

# use the levels to fetch columns and sum them 
o <- sapply(as.numeric(levels(ids)), function(x) { 
    rowSums(df[which(ids == x)])/df$length 
}) 

> o 
#   [,1]  [,2] [,3] 
# [1,] 3.000000 7.000000 1 
# [2,] 2.500000 4.500000 1 
# [3,] 2.333333 3.666667 1 
# [4,] 2.250000 3.250000 1 
# [5,] 2.200000 3.000000 1