英文:
Count how many times each string from a column appear (no exact match) in another column in R
问题
我的数据如下：
df <- data.frame(id = c("p3", "p5", "p8", "p9", "p10", "p11"), pedi = c("p1/p2", "p3/p4", "p3/p5", "(p3/p4)/p5", "p5/p8", "p4/p10"))
我正在尝试这样做：
id <- df$id
for (i in length(id)) {
df$id_in_pedi <- sum(grepl(i, df$pedi))
}
但它不起作用。我想要的结果是这样的：
df <- data.frame(id = c("p3", "p5", "p8", "p9", "p10", "p11"),
pedi = c("p1/p2", "p3/p4", "p3/p5", "(p3/p4)/p5", "p5/p8", "p4/p10"),
id_in_pedi = c(3, 3, 1, 0, 1, 0))
谢谢。
英文:
My data looks like this <br /> <br /> df <- data.frame(id = c("p3", "p5", "p8", "p9", "p10", "p11"), pedi = c("p1/p2", "p3/p4", "p3/p5", "(p3/p4)/p5", "p5/p8", "p4/p10"))<br /> <br /> I am trying this <br /> <br />
id <- df$id
for (i in length(id)) {
df$id_in_pedi <- sum(grepl(i, df$pedi))
}
<br /> But it does not work. The result I am looking for is this: <br />
df <- data.frame(id = c("p3", "p5", "p8", "p9", "p10", "p11"),
pedi = c("p1/p2", "p3/p4", "p3/p5", "(p3/p4)/p5", "p5/p8", "p4/p10"),
id_in_pedi = c(3,3,1,0,1,0))
<br /> Thanks
答案1
得分: 3
在tidyverse中：
library(tidyverse)
df %>%
mutate(id_in_pedi = str_count(toString(pedi), id))
在Base R中，使用sapply：
transform(df, id_in_pedi = colSums(sapply(id, grepl, pedi, USE.NAMES = FALSE)))
或者使用Vectorize：
colSums(Vectorize(grepl)(df$id, list(df$pedi)))
翻译完成。
英文:
In tidyverse:
library(tidyverse)
df %>%
mutate(id_in_pedi = str_count(toString(pedi), id))
id pedi id_in_pedi
1 p3 p1/p2 3
2 p5 p3/p4 3
3 p8 p3/p5 1
4 p9 (p3/p4)/p5 0
5 p10 p5/p8 1
6 p11 p4/p10 0
in Base R:
Using sapply:
transform(df, id_in_pedi = colSums(sapply(id, grepl, pedi, USE.NAMES = FALSE)))
id pedi id_in_pedi
1 p3 p1/p2 3
2 p5 p3/p4 3
3 p8 p3/p5 1
4 p9 (p3/p4)/p5 0
5 p10 p5/p8 1
6 p11 p4/p10 0
Using Vectorize:
colSums(Vectorize(grepl)(df$id, list(df$pedi)))
p3 p5 p8 p9 p10 p11
3 3 1 0 1 0
答案2
得分: 0
使用base R
table(factor(unlist(strsplit(df$pedi, "[/()]")), levels = df$id))
输出
p3 p5 p8 p9 p10 p11
3 3 1 0 1 0
英文:
Using base R
table(factor(unlist(strsplit(df$pedi, "[/()]")), levels = df$id))
-output
p3 p5 p8 p9 p10 p11
3 3 1 0 1 0
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