根据列名粘贴值

Question

我在问一个与@Axeman 的正确答案有关的问题，关于我的类似。下面是我的 dataframe 的 dput 和 output。

我正在尝试从 2 组变量（长度相同 N）中执行 paste 个字符串。

var1.x, var2.x, var1.y, var2.y 的字符串应该变成 var1, var2 。我正在寻找一个可以支持 varN.x, varN.y 成为 varN

的代码

我已经修改了@Axeman 的答案，但是在粘贴 N 个变量时它不会缩放。

df[,3:4] <- mapply(function(x, y) paste0(na.omit(c(x, y)), collapse = ''), 
                    as.character(df[,3:4]), as.character(df[,5:6]))
output <- df[1:4]

df <- structure(list(factor1 = structure(c(1L, 1L, 2L, 1L, 1L, 2L, 
2L, 1L), .Label = c("f1", "f2"), class = "factor"), factor2 = c(1L, 
2L, 1L, 3L, 4L, 2L, 3L, 5L), var1.x = structure(c(1L, 2L, NA, 
3L, 4L, 6L, 7L, 5L), .Label = c("a", "d", "g", "h", "j", "t", 
"y"), class = "factor"), var2.x = structure(c(NA, 1L, 2L, NA, 
1L, 2L, 2L, 2L), .Label = c("g", "s"), class = "factor"), var1.y = structure(c(4L, 
1L, NA, 2L, 2L, 2L, NA, 3L), .Label = c("f", "g", "h", "x"), class = "factor"), 
    var2.y = structure(c(4L, 2L, 2L, 1L, NA, 3L, 3L, 3L), .Label = c("a", 
    "g", "h", "t"), class = "factor")), .Names = c("factor1", 
"factor2", "var1.x", "var2.x", "var1.y", "var2.y"), class = "data.frame", row.names = c(NA, 
-8L))

output <- structure(list(factor1 = structure(c(1L, 1L, 2L, 1L, 1L, 2L, 
2L, 1L), .Label = c("f1", "f2"), class = "factor"), factor2 = c(1L, 
2L, 1L, 3L, 4L, 2L, 3L, 5L), var1 = structure(c(1L, 2L, NA, 3L, 
4L, 6L, 7L, 5L), .Label = c("ax", "df", "gg", "hg", "js", "tg", 
"y"), class = "factor"), var2 = structure(c(7L, 3L, 5L, 1L, 2L, 
6L, 6L, 4L), .Label = c("a", "g", "gg", "hh", "sg", "sh", "t"
), class = "factor")), .Names = c("factor1", "factor2", "var1", 
"var2"), class = "data.frame", row.names = c(NA, -8L))

Answer 1

一种通过 base R 实现的方法，

#make sure the columns you are pasting are characters
df[-c(1:2)] <- lapply(df[-c(1:2)], as.character)

#replace NA with '' to avoid pasting problems
df[is.na(df)] <- ''

#create a vector with unique column names 
ind <- unique(sub('\..*', '', names(df[-c(1:2)])))

#create a matrix matching each column name with ind, in order to use as index
m1 <- t(sapply(ind, grepl, names(df[-c(1:2)])))

#apply paste0 in columns based on index matrix m1.
df1 <- setNames(data.frame(sapply(seq(nrow(m1)), function(i)
                do.call(paste0, df[-c(1:2)][m1[i,]]))), paste0('Var', seq(nrow(m1))))

#bind it back to first two columns of original df and change '' to NA
df <- cbind(df[1:2], df1)
df[df == ''] <- NA

df
#  factor1 factor2 Var1 Var2
#1      f1       1   ax    t
#2      f1       2   df   gg
#3      f2       1 <NA>   sg
#4      f1       3   gg    a
#5      f1       4   hg    g
#6      f2       2   tg   sh
#7      f2       3    y   sh
#8      f1       5   jh   sh

根据列名粘贴值

pasting values based on column names

string

r

paste