根据所有其他列中是否存在 0/1 创建指标列答案

【问题标题】：Create indicator column based on presence of 0/1 in all other columns根据所有其他列中是否存在 0/1 创建指标列
【发布时间】：2021-08-18 15:53:14
【问题描述】：

我经常发现自己必须应用以下条件：我有一个包含多个二进制列的表，评级为 yes/no 或 0/1。我必须使用以下规则在计算中创建一个新的中间列：如果所有列都是“否”，那么新列是“否”，如果至少有一列有“是”，那么汇总列必须表示“是的”。我通常使用 case_when 执行此操作，并且效果很好（参见示例）。

library(tidyverse)

#create a table for reproducible example
set.seed(001)
carac1 <- round(runif(100),0)
carac2 <- round(runif(100),0)
carac3 <- round(runif(100),0)
data <- data.frame(carac1,carac2,carac3)

#apply case_when with complex condition
data <- data %>%
  mutate(carac_all = case_when(
    carac1 == 0 & carac2 == 0 & carac3 == 0 ~ "Always no",
    carac1 == 1 | carac2 == 1 | carac3 == 1 ~ "yes at least one time",
    TRUE ~ NA_character_))

这正是我想要的：

   carac1 carac2 carac3             carac_all
1        0      1      0 yes at least one time
2        0      0      0             Always no
3        1      0      1 yes at least one time
4        1      1      0 yes at least one time

（此示例使用数字 0/1，但有时使用字符是/否或其他类别变量，例如颜色......因此在这些情况下使用 >0 的技巧并不容易实现。）

问题是这段代码强制我在代码中输入每一列的名称。在我的最后一个文件中，我有 120 个连续的列要分析......有没有办法在一系列列上使用这种情况下的 case_when？我尝试了carac1:carac3 == 0，但它不起作用，无论如何我不明白如何表达“至少一列说是”。

感谢您的帮助。

TLDR：我想简化我当前使用的代码，这样我就不必在代码中输入每个变量的名称，而是输入一系列变量。

【问题讨论】：

标签： r dplyr

【解决方案1】：

3。更新：使用dplyr 1.0.4: if_any() and if_all()

data %>% 
    mutate(
        carac_all = case_when(
            if_all(contains("carac"), ~. < 1) ~ "Always no",
            if_any(contains("carac"), ~. >= 1) ~ "yes at least one time",
            TRUE ~ NA_character_))

2。更新：感谢 Martin Gal 的宝贵意见：

data %>% 
    mutate(carac_all = case_when(
        rowSums(across(carac1:carac3)) < 1 ~ "Always no",
        rowSums(across(carac1:carac3)) >=1 ~ "yes at least one time",
        TRUE ~ NA_character_))

更新：澄清后：

data %>% 
    mutate(sum_carac = carac1+carac2+carac3) %>% 
    mutate(carac_all = case_when(
        sum_carac < 1 ~ "Always no",
        sum_carac >=1 ~ "yes at least one time",
        TRUE ~ NA_character_)) %>% 
    select(-sum_carac)

  carac1 carac2 carac3             carac_all
1        0      1      0 yes at least one time
2        0      0      0             Always no
3        1      0      1 yes at least one time
4        1      1      0 yes at least one time
5        0      1      0 yes at least one time
6        1      0      1 yes at least one time
7        1      0      1 yes at least one time
8        1      0      0 yes at least one time
9        1      1      0 yes at least one time
10       0      1      1 yes at least one time
11       0      1      1 yes at least one time
12       0      1      0 yes at least one time
13       1      0      1 yes at least one time
14       0      0      1 yes at least one time
15       1      0      1 yes at least one time
16       0      0      0             Always no
17       1      1      1 yes at least one time
18       1      0      1 yes at least one time
19       0      0      1 yes at least one time
20       1      1      0 yes at least one time
21       1      1      0 yes at least one time
22       0      0      0             Always no
23       1      0      0 yes at least one time
24       0      0      1 yes at least one time
25       0      1      1 yes at least one time
26       0      0      1 yes at least one time
27       0      1      0 yes at least one time
28       0      0      0             Always no
29       1      0      0 yes at least one time
30       0      1      1 yes at least one time
31       0      1      0 yes at least one time
32       1      0      0 yes at least one time
33       0      0      0             Always no
34       0      1      1 yes at least one time
35       1      1      0 yes at least one time
36       1      1      1 yes at least one time
37       1      1      1 yes at least one time
38       0      1      1 yes at least one time
39       1      1      0 yes at least one time
40       0      1      0 yes at least one time
41       1      1      0 yes at least one time
42       1      1      1 yes at least one time
43       1      0      1 yes at least one time
44       1      0      0 yes at least one time
45       1      1      0 yes at least one time
46       1      0      0 yes at least one time
47       0      0      0             Always no
48       0      1      0 yes at least one time
49       1      0      0 yes at least one time
50       1      1      1 yes at least one time
51       0      1      1 yes at least one time
52       1      1      1 yes at least one time
53       0      0      0             Always no
54       0      0      1 yes at least one time
55       0      1      0 yes at least one time
56       0      0      0             Always no
57       0      1      0 yes at least one time
58       1      0      0 yes at least one time
59       1      0      0 yes at least one time
60       0      0      1 yes at least one time
61       1      0      1 yes at least one time
62       0      1      0 yes at least one time
63       0      0      0             Always no
64       0      1      1 yes at least one time
65       1      1      1 yes at least one time
66       0      0      0             Always no
67       0      0      0             Always no
68       1      0      1 yes at least one time
69       0      1      1 yes at least one time
70       1      0      0 yes at least one time
71       0      1      0 yes at least one time
72       1      1      0 yes at least one time
73       0      1      0 yes at least one time
74       0      0      0             Always no
75       0      0      0             Always no
76       1      1      0 yes at least one time
77       1      1      0 yes at least one time
78       0      1      0 yes at least one time
79       1      1      0 yes at least one time
80       1      1      1 yes at least one time
81       0      0      0             Always no
82       1      0      1 yes at least one time
83       0      1      1 yes at least one time
84       0      1      0 yes at least one time
85       1      1      0 yes at least one time
86       0      0      0             Always no
87       1      1      0 yes at least one time
88       0      1      0 yes at least one time
89       0      1      0 yes at least one time
90       0      1      0 yes at least one time
91       0      1      0 yes at least one time
92       0      0      0             Always no
93       1      0      1 yes at least one time
94       1      1      0 yes at least one time
95       1      0      1 yes at least one time
96       1      1      1 yes at least one time
97       0      0      0             Always no
98       0      1      0 yes at least one time
99       1      0      0 yes at least one time
100      1      1      1 yes at least one time

第一个答案： 我们可以使用来自dplyr 包的across

library(dplyr)
data %>% 
    mutate(across(starts_with("carac"), ~case_when(
        . == 0 ~ "Always no",
        . == 1 ~ "yes at least one time",
        TRUE ~ NA_character_), .names ="x_{.col}")) %>% 
        select(carac1:x_carac1)

 carac1 carac2 carac3             carac_all
1        0      1      0 yes at least one time
2        0      0      0             Always no
3        1      0      1 yes at least one time
4        1      1      0 yes at least one time
5        0      1      0 yes at least one time
6        1      0      1 yes at least one time
7        1      0      1 yes at least one time
8        1      0      0 yes at least one time
9        1      1      0 yes at least one time
10       0      1      1 yes at least one time
11       0      1      1 yes at least one time
12       0      1      0 yes at least one time
13       1      0      1 yes at least one time
14       0      0      1 yes at least one time
15       1      0      1 yes at least one time
16       0      0      0             Always no
17       1      1      1 yes at least one time
18       1      0      1 yes at least one time
19       0      0      1 yes at least one time
20       1      1      0 yes at least one time
21       1      1      0 yes at least one time
22       0      0      0             Always no
23       1      0      0 yes at least one time
24       0      0      1 yes at least one time
25       0      1      1 yes at least one time
26       0      0      1 yes at least one time
27       0      1      0 yes at least one time
28       0      0      0             Always no
29       1      0      0 yes at least one time
30       0      1      1 yes at least one time
31       0      1      0 yes at least one time
32       1      0      0 yes at least one time
33       0      0      0             Always no
34       0      1      1 yes at least one time
35       1      1      0 yes at least one time
36       1      1      1 yes at least one time
37       1      1      1 yes at least one time
38       0      1      1 yes at least one time
39       1      1      0 yes at least one time
40       0      1      0 yes at least one time
41       1      1      0 yes at least one time
42       1      1      1 yes at least one time
43       1      0      1 yes at least one time
44       1      0      0 yes at least one time
45       1      1      0 yes at least one time
46       1      0      0 yes at least one time
47       0      0      0             Always no
48       0      1      0 yes at least one time
49       1      0      0 yes at least one time
50       1      1      1 yes at least one time
51       0      1      1 yes at least one time
52       1      1      1 yes at least one time
53       0      0      0             Always no
54       0      0      1 yes at least one time
55       0      1      0 yes at least one time
56       0      0      0             Always no
57       0      1      0 yes at least one time
58       1      0      0 yes at least one time
59       1      0      0 yes at least one time
60       0      0      1 yes at least one time
61       1      0      1 yes at least one time
.........

【讨论】：

大师case_when:D
对于map，我的大脑很复杂，因此我尝试across :-)。
对你来说没有什么是复杂的:)
我不明白。您不会在代码中创建名为 carac_all 的列。您创建了三个名为“results_from_carac1”的列，直到 carac3。可见的“carac_all”是在我的示例中创建的，您可以通过选择刚刚创建的列来隐藏。直接用你的代码不行
您可以删除代码的第二行和最后一行，并将case_when 中的sum_carac 替换为rowSums(across(carac1:carac3))。

【解决方案2】：

使用rowSums的简单方法：

data$carac_all = ifelse(rowSums(data) > 0, "yes at least one time", "Always no")

   carac1 carac2 carac3             carac_all
1       0      1      0 yes at least one time
2       0      0      0             Always no
3       1      0      1 yes at least one time
4       1      1      0 yes at least one time
5       0      1      0 yes at least one time
6       1      0      1 yes at least one time
7       1      0      1 yes at least one time
8       1      0      0 yes at least one time
9       1      1      0 yes at least one time
10      0      1      1 yes at least one time
11      0      1      1 yes at least one time
12      0      1      0 yes at least one time
13      1      0      1 yes at least one time
14      0      0      1 yes at least one time
15      1      0      1 yes at least one time
16      0      0      0             Always no
17      1      1      1 yes at least one time
18      1      0      1 yes at least one time
19      0      0      1 yes at least one time
20      1      1      0 yes at least one time
21      1      1      0 yes at least one time
22      0      0      0             Always no
23      1      0      0 yes at least one time
24      0      0      1 yes at least one time
25      0      1      1 yes at least one time

【讨论】：

它可以工作（在 pivot_wider 的末尾没有“dr”）。在这种情况下，我很难真正理解 pivot_longer 中的转换，但在这个例子中它是有效的 ;) 是否需要使用 pivot ？

【解决方案3】：

您可以使用lookup 向量，其中1:3 是carac 的列索引。

lookup <- c("TRUE" = "yes at least one time", "FALSE" = "Always no")
data$carac_all <- lookup[paste0(apply(data[, 1:3], 1, sum) > 0)]

【讨论】：

【解决方案4】：

所有建议的解决方案都有效，但它们涉及求和计算：如果 sum >= 1，则 1 至少出现一次。我终于找到了我想象中的解决方案，即也可以使用分类变量。必须经过 if_any 和 if_all ：

library(tidyverse)
set.seed(001)
#create a table for reproducible example
carac1 <- round(runif(100),0)
carac2 <- round(runif(100),0)
carac3 <- round(runif(100),0)
data <- data.frame(carac1,carac2,carac3)

data <- data %>%
  mutate(carac_all = case_when(
    if_any(carac1:carac3, ~.x == "1") == T ~ "yes at least one time",
    if_all(carac1:carac3, ~.x == "0") == T ~ "always no",
    TRUE ~ NA_character_))

非常感谢大家

【讨论】：