我有一个 df.如果“info”出现在一行中,我想要一个函数来获取列名称并将其与单元格中的信息粘贴在一起,并在它们之间使用“=”。我已经尝试过下面的函数,该函数有效,只是它不引用右列中的列名
df <- data.frame('criteria1' = c('info','1', 'info', '', 'info'), "criteria2" = c('y','3', '', 'info', ''), "criteria3" = c('y','7', '', 'info', 'info')) df[] <- sapply(df, function(x) ifelse(x == 'info', paste(colnames(df)[x], ' = ', x),''))
我的预期输出是这样的(列名是否被删除并不重要,重要的是单元格中的信息)
df_exp <- data.frame('criteria1' = c('criteria1= info','', 'criteria1=info', '', 'criteria1 =info'), "criteria2" = c('','', '', 'criteria2 = info', ''), "criteria3" = c('','', '', 'criteria3 = info', 'criteria3 = info'))
最佳答案
我会选择列号,1:ncol(df)
(更快)或seq(df)
。我用的是前者。
df <- sapply(1:ncol(df), function(x)
ifelse(df[[x]] == 'info', paste(colnames(df)[x], ' = ', df[[x]]),''))
df
# [,1] [,2] [,3]
# [1,] "criteria1 = info" "" ""
# [2,] "" "" ""
# [3,] "criteria1 = info" "" ""
# [4,] "" "criteria2 = info" "criteria3 = info"
# [5,] "criteria1 = info" "" "criteria3 = info"
使用stack/unstack
的另一种好方法:
r <- grep("info", tmp$values)
tmp <- stack(df)
tmp[r, 1] <- apply(tmp[r, 2:1], 1, paste, collapse="=")
tmp[-r, 1] <- "" ## in case you want non-"info" cells cleared
df <- unstack(tmp)
df
# criteria1 criteria2 criteria3
# 1 criteria1=info
# 2
# 3 criteria1=info
# 4 criteria2=info criteria3=info
# 5 criteria1=info criteria3=info
关于r - 如何在函数中引用列名,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/62638554/