[英]How to create a new categorical variable based on the location of first zero in a column in a long format data using R?
数据集以长格式布局，没有缺失值，有 5 列，第 1 列是“id”，第 2 列是“gender”，第 3 列是二进制变量“competency”，第 4 个变量是薪水，第 5 列是时间。
现在，我想根据以下规则通过id创建一个新的分类变量“yeel”（年终评估级别），
1-如果每个id内的“competency”变量的值都等于1，那么“yeel”的值应该是“unqualified”；
2-如果变量“competency”中第一个零的位置落在“time”变量的[1,5]范围内，则“yeel”的值应该是“entry level”；
3-如果变量“competency”中第一个零的位置落在“time”变量的[6,8]范围内，则“yeel”的值应该是“intermediate level”；
4-如果变量“competency”中第一个零的位置落在“time”变量的[9,12]范围内，则“yeel”的值应该是“senior level”；
5-如果每个id中“competency”变量的所有值都等于0，那么“yeel”的值应该是“undefined”；
例如，预期的数据集应该如下所示
structure(list(id = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5,
5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7,
7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9,
9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10,
10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11,
11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13,
13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14,
14, 14, 14, 14, 14), gender = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), competency = c(1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0,
0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1,
1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1,
1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), salary = c(15000,
15500, 15800, 16000, 16200, 16400, 16600, 16800, 17000, 17300,
17500, 17500, 15000, 16000, 17000, 18000, 19000, 20000, 21000,
22000, 23000, 24000, 25000, 26000, 15000, 16000, 18000, 20000,
22000, 24000, 26000, 28000, 30000, 32000, 34000, 36000, 15000,
16000, 17000, 17500, 18000, 18500, 19000, 19500, 20000, 20500,
21000, 21500, 15000, 15300, 15600, 15900, 16200, 16500, 16800,
17100, 17400, 17700, 18000, 18300, 15000, 15500, 15800, 16000,
16200, 16400, 16600, 16800, 17000, 17300, 17500, 17500, 15000,
16000, 17000, 18000, 19000, 20000, 21000, 22000, 23000, 24000,
25000, 26000, 15000, 16000, 18000, 20000, 22000, 24000, 26000,
28000, 30000, 32000, 34000, 36000, 15000, 16000, 17000, 17500,
18000, 18500, 19000, 19500, 20000, 20500, 21000, 21500, 15000,
15300, 15600, 15900, 16200, 16500, 16800, 17100, 17400, 17700,
18000, 18300, 15000, 15500, 15800, 16000, 16200, 16400, 16600,
16800, 17000, 17300, 17500, 17500, 15000, 16000, 17000, 18000,
19000, 20000, 21000, 22000, 23000, 24000, 25000, 26000, 15000,
16000, 18000, 20000, 22000, 24000, 26000, 28000, 30000, 32000,
34000, 36000, 36000, 36000, 36000, 36000, 36000, 36000, 36000,
36000, 36000, 36000, 36000, 36000), time = c(1, 2, 3, 4, 5, 6,
7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1,
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4,
5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,
12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6,
7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1,
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4,
5, 6, 7, 8, 9, 10, 11, 12), yeel = c("unqualified", "unqualified",
"unqualified", "unqualified", "unqualified", "unqualified", "unqualified",
"unqualified", "unqualified", "unqualified", "unqualified", "unqualified",
"entry level", "entry level", "entry level", "entry level", "entry level",
"entry level", "entry level", "entry level", "entry level", "entry level",
"entry level", "entry level", "entry level", "entry level", "entry level",
"entry level", "entry level", "entry level", "entry level", "entry level",
"entry level", "entry level", "entry level", "entry level", "entry level",
"entry level", "entry level", "entry level", "entry level", "entry level",
"entry level", "entry level", "entry level", "entry level", "entry level",
"entry level", "entry level", "entry level", "entry level", "entry level",
"entry level", "intermediate level", "intermediate level", "intermediate level",
"intermediate level", "intermediate level", "intermediate level",
"intermediate level", "intermediate level", "intermediate level",
"intermediate level", "intermediate level", "intermediate level",
"intermediate level", "intermediate level", "intermediate level",
"intermediate level", "intermediate level", "intermediate level",
"intermediate level", "intermediate level", "intermediate level",
"intermediate level", "intermediate level", "intermediate level",
"intermediate level", "intermediate level", "intermediate level",
"intermediate level", "intermediate level", "intermediate level",
"intermediate level", "intermediate level", "intermediate level",
"intermediate level", "intermediate level", "intermediate level",
"intermediate level", "intermediate level", "intermediate level",
"intermediate level", "intermediate level", "intermediate level",
"intermediate level", "senior level", "senior level", "senior level",
"senior level", "senior level", "senior level", "senior level",
"senior level", "senior level", "senior level", "senior level",
"senior level", "senior level", "senior level", "senior level",
"senior level", "senior level", "senior level", "senior level",
"senior level", "senior level", "senior level", "senior level",
"senior level", "senior level", "senior level", "senior level",
"senior level", "senior level", "senior level", "senior level",
"senior level", "senior level", "senior level", "senior level",
"senior level", "senior level", "senior level", "senior level",
"senior level", "senior level", "senior level", "senior level",
"senior level", "senior level", "senior level", "senior level",
"senior level", "senior level", "senior level", "senior level",
"senior level", "senior level", "senior level", "senior level",
"senior level", "senior level", "senior level", "senior level",
"senior level", "undefined", "undefined", "undefined", "undefined",
"undefined", "undefined", "undefined", "undefined", "undefined",
"undefined", "undefined", "undefined")), row.names = c(NA, 168L
), class = "data.frame")->df_new
原始数据集结构如下，
structure(list(id = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5,
5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7,
7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9,
9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10,
10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11,
11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13,
13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14,
14, 14, 14, 14, 14), gender = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), competency = c(1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0,
0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1,
1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1,
1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), salary = c(15000,
15500, 15800, 16000, 16200, 16400, 16600, 16800, 17000, 17300,
17500, 17500, 15000, 16000, 17000, 18000, 19000, 20000, 21000,
22000, 23000, 24000, 25000, 26000, 15000, 16000, 18000, 20000,
22000, 24000, 26000, 28000, 30000, 32000, 34000, 36000, 15000,
16000, 17000, 17500, 18000, 18500, 19000, 19500, 20000, 20500,
21000, 21500, 15000, 15300, 15600, 15900, 16200, 16500, 16800,
17100, 17400, 17700, 18000, 18300, 15000, 15500, 15800, 16000,
16200, 16400, 16600, 16800, 17000, 17300, 17500, 17500, 15000,
16000, 17000, 18000, 19000, 20000, 21000, 22000, 23000, 24000,
25000, 26000, 15000, 16000, 18000, 20000, 22000, 24000, 26000,
28000, 30000, 32000, 34000, 36000, 15000, 16000, 17000, 17500,
18000, 18500, 19000, 19500, 20000, 20500, 21000, 21500, 15000,
15300, 15600, 15900, 16200, 16500, 16800, 17100, 17400, 17700,
18000, 18300, 15000, 15500, 15800, 16000, 16200, 16400, 16600,
16800, 17000, 17300, 17500, 17500, 15000, 16000, 17000, 18000,
19000, 20000, 21000, 22000, 23000, 24000, 25000, 26000, 15000,
16000, 18000, 20000, 22000, 24000, 26000, 28000, 30000, 32000,
34000, 36000, 36000, 36000, 36000, 36000, 36000, 36000, 36000,
36000, 36000, 36000, 36000, 36000), time = c(1, 2, 3, 4, 5, 6,
7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1,
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4,
5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,
12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6,
7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1,
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4,
5, 6, 7, 8, 9, 10, 11, 12)), row.names = c(NA, 168L), class = "data.frame")->df
Thank you for your help!
您可以使用case_when为每个id传递多个条件。
library(dplyr)
df %>%
group_by(id) %>%
mutate(first_zero_loc = time[match(0, competency)],
yeel = case_when(all(competency == 1) ~ "unqualfied",
all(competency == 0) ~ "undefined",
between(first_zero_loc, 1, 5) ~ "entry level",
between(first_zero_loc, 6, 8) ~ "intermediate level",
between(first_zero_loc, 9, 12) ~ "competency level",
)) %>%
ungroup %>%
select(-first_zero_loc)
当数据为长格式时，如何在 R 中创建具有条件的新变量？
[英]How to create a new variable with condition in R when data is long format?
在不拆分我的数据的情况下，我如何在 R 中使用 function 创建一个新的分类变量
[英]Without splitting my data, how can i create a new categorical variable using function in R
R：基于连续变量从分类变量创建新的分类变量
[英]R: create a new categorical variable from a categorical variable based on a continuous variable
如何基于此特定逻辑在我的 R 数据框中创建新变量（列）？
[英]How to create a new variable (column) in my R data frame based on this specific logic?
如何根据 R 中的行值组合创建新变量（列）？
[英]How to create a new variable (column) based on a combination of row values in R?
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