如何在特定列中删除零值的行？

Question

假设我的数据框是这样的

col1    col2    col3    col4    col5    col6    col7
------------------------------------------------------
  0       0       0       0     16,75   17,50   18,08
 18      24      24      24     19,83   20,47    0,00
 18      24      24      24      0,00   21,17   20,73
  0      22       0       0     18,67   18,90   21,23
 18      24      24      24      0,00   20,42   21,17
 18      24      24      24     20,52   21,17   21,92

我想在列时删除行col5,col6并col7包含0.最后,数据框的形状应如下所示:

col1    col2    col3    col4    col5    col6    col7
-----------------------------------------------------
  0      22       0       0     18,67   18,90   21,23
 18      24      24      24     20,52   21,17   21,92

Answer 1

我们可以用 filter_at

library(tidyverse)
df1 %>% 
   filter_at(vars(col5, col6, col7), all_vars(. != '0,00'))

Answer 2

基础R解决方案:

sapply找到不等于0的记录,如果整行只包含TRUE值和我们在data.frame中选择的值,则围绕它应用测试.

df1[apply(sapply(df1[, 5:7], function(x) x != 0), 1, all), ]

  col1 col2 col3 col4  col5  col6  col7
1    0    0    0    0 16.75 17.50 18.08
4    0   22    0    0 18.67 18.90 21.23
6   18   24   24   24 20.52 21.17 21.92

数据(我用dec ="读取你的数据",所以所有数据都被读作数字):

df1 <- structure(list(col1 = c(0L, 18L, 18L, 0L, 18L, 18L), col2 = c(0L, 
24L, 24L, 22L, 24L, 24L), col3 = c(0L, 24L, 24L, 0L, 24L, 24L
), col4 = c(0L, 24L, 24L, 0L, 24L, 24L), col5 = c(16.75, 19.83, 
0, 18.67, 0, 20.52), col6 = c(17.5, 20.47, 21.17, 18.9, 20.42, 
21.17), col7 = c(18.08, 0, 20.73, 21.23, 21.17, 21.92)), class = "data.frame", row.names = c(NA, 
-6L))