英文:
How to find the leftmost column of a data frame with all zeros in columns to the right with dplyr
问题
以下是您要翻译的内容：
I have a data frame like this:
df <- data.frame(v1=c(0,1), v2=c(0,1),v3=c(0,1), v4=c(0,1) ) %>%
expand(v1,v2,v3,v4)
df
# A tibble: 16 x 4
var1 var2 var3 var4
<dbl> <dbl> <dbl> <dbl>
1 0 0 0 0
2 0 0 0 1
3 0 0 1 0
4 0 0 1 1
5 0 1 0 0
6 0 1 0 1
7 0 1 1 0
8 0 1 1 1
9 1 0 0 0
10 1 0 0 1
11 1 0 1 0
12 1 0 1 1
13 1 1 0 0
14 1 1 0 1
15 1 1 1 0
16 1 1 1 1
For each row, I would like to identify the leftmost column that contains a zero, such that all columns to the right also contain 0s. The expected output is as follows:
# A tibble: 16 x 5
var1 var2 var3 var4 result
<dbl> <dbl> <dbl> <dbl> <chr>
1 0 0 0 0 var1
2 0 0 0 1 NA
3 0 0 1 0 var4
4 0 0 1 1 NA
5 0 1 0 0 var3
6 0 1 0 1 NA
7 0 1 1 0 var4
8 0 1 1 1 NA
9 1 0 0 0 var2
10 1 0 0 1 NA
11 1 0 1 0 var4
12 1 0 1 1 NA
13 1 1 0 0 var3
14 1 1 0 1 NA
15 1 1 1 0 var4
16 1 1 1 1 NA
If possible, I would prefer a tidyverse solution.
<details>
<summary>英文:</summary>
I have a data frame like this:
df <- data.frame(v1=c(0,1), v2=c(0,1),v3=c(0,1), v4=c(0,1) ) %>%
expand(v1,v2,v3,v4)
df
A tibble: 16 x 4
var1 var2 var3 var4
<dbl> <dbl> <dbl> <dbl>
1 0 0 0 0
2 0 0 0 1
3 0 0 1 0
4 0 0 1 1
5 0 1 0 0
6 0 1 0 1
7 0 1 1 0
8 0 1 1 1
9 1 0 0 0
10 1 0 0 1
11 1 0 1 0
12 1 0 1 1
13 1 1 0 0
14 1 1 0 1
15 1 1 1 0
16 1 1 1 1
For each row, I would like to identify the leftmost column that contains a zero, such that all columns to the right also contain 0s. The expected output is as follows:
A tibble: 16 x 5
var1 var2 var3 var4 result
<dbl> <dbl> <dbl> <dbl> <chr>
1 0 0 0 0 var1
2 0 0 0 1 NA
3 0 0 1 0 var4
4 0 0 1 1 NA
5 0 1 0 0 var3
6 0 1 0 1 NA
7 0 1 1 0 var4
8 0 1 1 1 NA
9 1 0 0 0 var2
10 1 0 0 1 NA
11 1 0 1 0 var4
12 1 0 1 1 NA
13 1 1 0 0 var3
14 1 1 0 1 NA
15 1 1 1 0 var4
16 1 1 1 1 NA
If possible, I would prefer a tidyverse solution.
</details>
# 答案1
**得分**: 3
这是您提供的代码的中文翻译部分：
```R
我对tidyverse解决方案一无所知，但这里有一个基本的R答案：
chk <- simplify2array(rev(Reduce(`+`, rev(df), accumulate=TRUE))) == 0
df$name <- names(df)[max.col(chk, "first")]
df$name[rowSums(chk) == 0] <- NA
df
这段代码的作用是将数据框df中的列进行一些操作，然后将结果存储在df的"name"列中。
英文:
I've no idea about a tidyverse solution, but here's a base R answer:
chk <- simplify2array(rev(Reduce(`+`, rev(df), accumulate=TRUE))) == 0
df$name <- names(df)[max.col(chk, "first")]
df$name[rowSums(chk) == 0] <- NA
df
### A tibble: 16 × 5
## v1 v2 v3 v4 name
## <dbl> <dbl> <dbl> <dbl> <chr>
## 1 0 0 0 0 v1
## 2 0 0 0 1 NA
## 3 0 0 1 0 v4
## 4 0 0 1 1 NA
## 5 0 1 0 0 v3
## 6 0 1 0 1 NA
## 7 0 1 1 0 v4
## 8 0 1 1 1 NA
## 9 1 0 0 0 v2
##10 1 0 0 1 NA
##11 1 0 1 0 v4
##12 1 0 1 1 NA
##13 1 1 0 0 v3
##14 1 1 0 1 NA
##15 1 1 1 0 v4
##16 1 1 1 1 NA
答案2
得分: 3
这个解决方案将每一行的数值传递给一个辅助函数，使用 dplyr::c_across()；然后辅助函数使用 rle() 来返回最后一组零值的起始位置，如果最后一个值是零，否则返回 NA。
library(dplyr)
find_zero <- function(x, cols) {
x_rle <- rle(x)
if (tail(x_rle$values, 1) != 0) NA
else rev(cols)[[tail(x_rle$lengths, 1)]]
}
df %>%
rowwise() %>%
mutate(result = find_zero(c_across(v1:v4), names(.))) %>%
ungroup()
# A tibble: 16 × 5
v1 v2 v3 v4 result
<dbl> <dbl> <dbl> <dbl> <chr>
1 0 0 0 0 v1
2 0 0 0 1 <NA>
3 0 0 1 0 v4
4 0 0 1 1 <NA>
5 0 1 0 0 v3
6 0 1 0 1 <NA>
7 0 1 1 0 v4
8 0 1 1 1 <NA>
9 1 0 0 0 v2
10 1 0 0 1 <NA>
11 1 0 1 0 v4
12 1 0 1 1 <NA>
13 1 1 0 0 v3
14 1 1 0 1 <NA>
15 1 1 1 0 v4
16 1 1 1 1 <NA>
英文:
This solution passes each row of values to a helper function using dplyr::c_across(); the helper function then uses rle() to return the start of the last run of zeros, if the last value is zero, and NA otherwise.
library(dplyr)
find_zero <- function(x, cols) {
x_rle <- rle(x)
if (tail(x_rle$values, 1) != 0) NA
else rev(cols)[[tail(x_rle$lengths, 1)]]
}
df %>%
rowwise() %>%
mutate(result = find_zero(c_across(v1:v4), names(.))) %>%
ungroup()
# A tibble: 16 × 5
v1 v2 v3 v4 result
<dbl> <dbl> <dbl> <dbl> <chr>
1 0 0 0 0 v1
2 0 0 0 1 <NA>
3 0 0 1 0 v4
4 0 0 1 1 <NA>
5 0 1 0 0 v3
6 0 1 0 1 <NA>
7 0 1 1 0 v4
8 0 1 1 1 <NA>
9 1 0 0 0 v2
10 1 0 0 1 <NA>
11 1 0 1 0 v4
12 1 0 1 1 <NA>
13 1 1 0 0 v3
14 1 1 0 1 <NA>
15 1 1 1 0 v4
16 1 1 1 1 <NA>
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