r-更改一个新列，并根据列名中的字符串从现有列中粘贴值

Question

我有这个数据帧：

df <- structure(list(number = 1:3, a_1 = c(1L, 4L, 7L), a_2 = c(2L, 
5L, 8L), a_3 = c(3L, 6L, 9L)), class = "data.frame", row.names = c(NA, 
-3L))
number a_1 a_2 a_3
1      1   1   2   3
2      2   4   5   6
3      3   7   8   9

我想对一个new_col进行变异，并用值填充它，条件是列number和列名的字符串匹配。

期望输出：

number   a_1   a_2   a_3 new_col
<int> <int> <int> <int>   <int>
1      1     1     2     3       1
2      2     4     5     6       5
3      3     7     8     9       9

我试过str_extract、str_detect。。。但我做不到！

Answer 1

或者使用一些pivot_*:

library(tidyr)
library(dplyr)
df %>% 
pivot_longer(-number, names_pattern = "a_(\d+)") %>% 
group_by(number) %>% 
mutate(new_col = value[name == number]) %>% 
pivot_wider(names_from = name, names_prefix = "a_") %>% 
ungroup()

返回

# A tibble: 3 x 5
number new_col   a_1   a_2   a_3
<int>   <int> <int> <int> <int>
1      1       1     1     2     3
2      2       5     4     5     6
3      3       9     7     8     9

Answer 2

我们可以在paste之后使用get，使用带有"number"的"a_">

library(dplyr)
library(stringr)
df %>% 
rowwise %>%
mutate(new_col = get(str_c('a_', number))) %>%
ungroup

-输出

# A tibble: 3 x 5
number   a_1   a_2   a_3 new_col
<int> <int> <int> <int>   <int>
1      1     1     2     3       1
2      2     4     5     6       5
3      3     7     8     9       9

---

使用row/column索引的矢量化选项可能更好

df$newcol <- df[-1][cbind(seq_len(nrow(df)),
match(paste0("a_", df$number), names(df)[-1]))]

Answer 3

或者这个：

library(dplyr)
df %>%
rowwise() %>%
mutate(new_col = c_across(starts_with("a"))[grepl(number, names(df[-1]))])
# A tibble: 3 x 5
# Rowwise: 
number   a_1   a_2   a_3 new_col
<int> <int> <int> <int>   <int>
1      1     1     2     3       1
2      2     4     5     6       5
3      3     7     8     9       9