我在类似的question上提出与@Axeman正确答案有关的问题。dput
的{{1}}和dataframe
的{{1}}位于下方。
我正在尝试从2组变量(长度相同的N)中执行output
个字符串。
paste
的字符串应为var1.x, var2.x, var1.y, var2.y
。我正在寻找可以支持var1, var2
成为varN.x, varN.y
我已经调整了@ Axeman的答案,但是在粘贴N个变量时它不会缩放。
varN
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))
答案 0 :(得分:1)
通过基地R的一种方式,
#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