Question

我想按规则对级别名称进行修改，但我在下面遇到问题：

我的数据;最初的df变量是类矩阵我改变了它

df <- data.frame(x = c("P27C", "P31B", "P12E", "P3E", "P7A", "P7D", "P2A", "P7D",
"P34", "P10C"),
             y = rnorm(10), stringsAsFactors = F)

s<-c("P27CvsP31B","P27CvsP3C","P27CvsP3E","P27CvsP6B","P27CvsP7A","P27CvsP7C",   
"P27DvsP27E","P27DvsP2B","P27DvsP31A","P27DvsP31B","P27DvsP3D","P27DvsP7D",
"P27EvsP2A","P27EvsP2B","P27EvsP2E","P27EvsP2F","P27EvsP2G","P27EvsP34", 
"P7AvsP7H","P7BvsP7D","P7CvsP7G","P7DvsP7E","P7DvsP7F","P7DvsP7G","P7DvsP7H") 

df

df$z <- lapply(df$x, grep, s, value = T)

# gives you the matches but empty slots for a missing value like "P12E"

df

for (r in 1:nrow(df)) {
    if (length(df$z[[r]]) == 0) {
        df$z[[r]] <- df$x[[r]]
        }
    else {
        df$z[[r]] <- df$z[[r]]
    }
}

# restores the original name of unmatched values

df$z 

#Rename but in list format!!!

我想要的输出是：

     x           y                                                                 z
1  P27C  2.22354499 "P27CvsP31B, P27CvsP3C, P27CvsP3E, P27CvsP6B, P27CvsP7A, P27CvsP7C"
2  P31B  0.89197064                                            "P27CvsP31B, P27DvsP31B"
3  P12E -0.02313754                                                              "P12E"
4   P3E  0.69916446                                                         "P27CvsP3E"
5   P7A -0.44895512                                               "P27CvsP7A, P7AvsP7H"
6   P7D  1.77619979       "P27DvsP7D, P7BvsP7D, P7DvsP7E, P7DvsP7F, P7DvsP7G, P7DvsP7H"
7   P2A -0.18261732                                                         "P27EvsP2A"
8   P7D  0.12025524       "P27DvsP7D, P7BvsP7D, P7DvsP7E, P7DvsP7F, P7DvsP7G, P7DvsP7H"
9   P34 -0.13434265                                                         "P27EvsP34"
10 P10C  0.19971201                                                              "P10C"

由于

Answer 1

嵌套sapply看起来有点难看。它遍历x的{{1}}列，并与您的向量df匹配所有条目，创建匹配结果的列表。第二个s循环遍历该列表并粘贴所有条目。如果没有匹配，则它返回一个空单元格，我们通过在其位置替换sapply条目来处理它。

df$x

修改

基于@ akrun的建议，df$z <- sapply(sapply(df$x, function(i) s[grepl(i, s)]), paste, collapse = ',') df$z[df$z == ''] <- df$x[df$z == ''] df # x y z #1 P27C -0.95290496 P27CvsP31B,P27CvsP3C,P27CvsP3E,P27CvsP6B,P27CvsP7A,P27CvsP7C #2 P31B 1.62237939 P27CvsP31B,P27DvsP31B #3 P12E 2.60014202 P12E #4 P3E 0.13964851 P27CvsP3E #5 P7A -1.35071967 P27CvsP7A,P7AvsP7H #6 P7D 0.79893102 P27DvsP7D,P7BvsP7D,P7DvsP7E,P7DvsP7F,P7DvsP7G,P7DvsP7H #7 P2A -1.55499584 P27EvsP2A #8 P7D 0.46372006 P27DvsP7D,P7BvsP7D,P7DvsP7E,P7DvsP7F,P7DvsP7G,P7DvsP7H #9 P34 0.05242956 P27EvsP34 #10 P10C -0.20203180 P10C的选项将是，

data.table

在名称级别创建长字符串

1 个答案: