How to subset dataframe on lowercase values in multiple columns

Question

I have a dataframe like this:

set.seed(12)
df <- data.frame(
  v1 = sample(LETTERS, 10),
  v2 = sample(LETTERS, 10),
  v3 = sample(LETTERS, 10),
  v4 = c(sample(LETTERS, 8), sample(letters, 2)),
  v5 = c(sample(letters, 1), sample(LETTERS, 7), sample(letters, 2))
    )
df
   v1 v2 v3 v4 v5
1   B  K  F  G  p
2   U  U  T  W  N
3   W  J  C  V  Y
4   G  I  Q  S  E
5   D  F  E  N  T
6   A  X  Z  T  C
7   V  Y  K  X  I
8   M  Z  D  Q  A
9   Y  L  H  k  d
10  R  B  L  j  t

I want to subset dfon those rows that contain a lowercase value in any of df's columns. It can be done like this:

df1 <- df[grepl("[a-z]", df$v1) | grepl("[a-z]", df$v2) | grepl("[a-z]", df$v3) |
          grepl("[a-z]", df$v4) | grepl("[a-z]", df$v5), ]
df1
   v1 v2 v3 v4 v5
1   B  K  F  G  p
9   Y  L  H  k  d
10  R  B  L  j  t

但这是不经济的，如果您有很多（更多）列，并且容易出错。有没有更干净、更简单、更经济的方法，最好是在基 R 中？

Answer 1

df[rowSums(sapply(df, function(x) x %in% letters)) > 0,]
#OR
df[apply(df == sapply(df, tolower), 1, any),]
#   v1 v2 v3 v4 v5
#1   B  L  L  M  e
#9   R  N  D  t  t
#10  F  X  M  h  x

Answer 2

一种选择是grepl在每一列上应用lapply以创建一个list逻辑vectors 和Reduce它|

df[Reduce(`|`, lapply(df, grepl, pattern = "[a-z]")),]
#   v1 v2 v3 v4 v5
#1   B  L  L  M  e
#9   R  N  D  t  t
#10  F  X  M  h  x

或使用 filter_all

library(dplyr)
library(stringr)
df %>% 
    filter_all(any_vars(str_detect(., "[a-z]")))
#  v1 v2 v3 v4 v5
#1  B  L  L  M  e
#2  R  N  D  t  t
#3  F  X  M  h  x

Answer 3

您可以使用paste每行，然后使用grepl.

df[grepl("[a-z]", apply(df, 1, paste, collapse="")),]
#   v1 v2 v3 v4 v5
#1   B  L  L  M  e
#9   R  N  D  t  t
#10  F  X  M  h  x

或者另一个选择是do.call

df[grepl("[a-z]", do.call(paste, df)),]