使用两个标准对 R 中的数据框进行子集化，其中之一是正则表达式 [英] subsetting dataframe in R using two criteria, one of them is regular expression

问题描述

我有一个类似这样的数据集:

I have a dataset something like this:

col_a col_b    col_c
1     abc_boy  1
2     abc_boy  2
1     abc_girl 1
2     abc_girl 2

我只需要根据col_b和col_c取第一行，然后把col_c中的valye改一下就行了像这样:

I need to pick up the first row only based on col_b and col_c, and then change the valye in col_c, which is something like this:

df[grep("_boy$",df[,"col_b"]) &df[,"col_c"]=="1","col_c"] <- "是"

但是上面的代码是不行的，因为第一个条件和第二个条件不是来自同一个集合.

But the code above is not OK, since the first criteria and the second criteria do not originate from the same set.

我可以通过使用显式循环以愚蠢的方式完成它，或者执行两层"子集，如下所示:

I can do it in a dumb way by using a explicit loop, or do a "two-tier" subsetting, something like this:

df.a <- df[grep("_boy$",df[,"col_b"]),]              #1
df.b <- df[grep("_boy$",df[,"col_b"],invert=TRUE),]  #2
df.a <- df.a[df.a[,"col_c"]=="1","col_c"] <- "yes"   #3
df.a <- df.a[df.a[,"col_c"]=="2","col_c"] <- "no"    #4
df <- rbind(df.a,df.b)                               #5

但我不想这样做，谁能告诉我如何合并"#1 和#3?谢谢.

But I prefer not to, can anyone enlighten me how to "merge" #1 and #3? Thanks.

推荐答案

尝试 grepl 而不是 grep.grepl 返回一个逻辑向量(x 的每个元素是否匹配)，可以与逻辑运算符组合.

Try grepl instead of grep. grepl returns a logical vector (match or not for each element of x), which can be combined with logical operators.

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