r - 使用 base R 根据另一个数据集中的值索引替换数据集中的值

Question

structure(list(ID = c(123, 5345, 234, 453, 3656, 345), diagnosis_1 = c("B657", 
"B658", "B659", "B660", "B661", "B662"), diagnosis_2 = c("F8827", 
"G432", NA, "B657", NA, "H8940"), diagnosis_3 = c(NA, "B657", 
NA, NA, NA, "G432"), diagnosis_4 = c(NA, NA, NA, NA, NA, "B657"
), diagnosis_5 = c(NA, NA, NA, NA, NA, NA), diagnosis_6 = c(NA, 
NA, NA, NA, NA, NA), diagnosis_7 = c(NA, NA, NA, NA, NA, NA), 
    diagnosis_8 = c(NA, NA, NA, NA, NA, NA), diagnosis_9 = c(NA, 
    NA, NA, NA, NA, NA), diagnosis_10 = c(NA, NA, NA, NA, NA, 
    NA), diagnosis_11 = c(NA, NA, NA, NA, NA, NA), diagnosis_12 = c(NA, 
    NA, NA, NA, NA, NA), diagnosis_13 = c(NA, NA, NA, NA, NA, 
    NA), age = c(54, 65, 23, 22, 33, 77)), row.names = c(NA, 
-6L), class = "data.frame")

我想用该索引中的值替换诊断列中的值：

B657    1
B658    2
B659    3
B660    4
B661    5
B662    1
F8827   3
G432    3
H8940   4

实际上，该表有数千行，并且我处理具有可变数量的诊断列的其他表，因此与列数无关的解决方案将是理想的。该索引也长达数百个条目。

如果索引表是这样划分的：

1 B657, B662
2 B658
3 B659, F8827, G432 
4 B660 H8940    
5 B661

这会对它的编码方式产生影响吗？

所需的输出如下所示：

非常感谢

标签： rdataframeindexingdplyrrecode

Answer 1

你可以试试这个

df_replace <- read.table(text="B657    1
B658    2
B659    3
B660    4
B661    5
B662    1
F8827   3
G432    3
H8940   4", stringsAsFactors = F)

vec_repl <-  as.character(df_replace$V2)
names(vec_repl) <- df_replace$V1

library(tidyverse)
df %>% 
  mutate_at(vars(starts_with("diag")), ~str_replace_all(., vec_repl))
    ID diagnosis_1 diagnosis_2 diagnosis_3 diagnosis_4 diagnosis_5 diagnosis_6 diagnosis_7 diagnosis_8 diagnosis_9 diagnosis_10 diagnosis_11 diagnosis_12 diagnosis_13
1  123           1           3        <NA>        <NA>        <NA>        <NA>        <NA>        <NA>        <NA>         <NA>         <NA>         <NA>         <NA>
2 5345           2           3           1        <NA>        <NA>        <NA>        <NA>        <NA>        <NA>         <NA>         <NA>         <NA>         <NA>
3  234           3        <NA>        <NA>        <NA>        <NA>        <NA>        <NA>        <NA>        <NA>         <NA>         <NA>         <NA>         <NA>
4  453           4           1        <NA>        <NA>        <NA>        <NA>        <NA>        <NA>        <NA>         <NA>         <NA>         <NA>         <NA>
5 3656           5        <NA>        <NA>        <NA>        <NA>        <NA>        <NA>        <NA>        <NA>         <NA>         <NA>         <NA>         <NA>
6  345           1           4           3           1        <NA>        <NA>        <NA>        <NA>        <NA>         <NA>         <NA>         <NA>         <NA>
  age
1  54
2  65
3  23
4  22
5  33
6  77

在基础中，您可以尝试使用R附加包stingr

df2 <- df
# use -c(1,ncol(df)) to select only columns where to replace values. 
df2[,-c(1,ncol(df))] <- lapply(df[,-c(1,ncol(df))], function(x) str_replace_all(x, vec_repl))
head(df2)
    ID diagnosis_1 diagnosis_2 diagnosis_3 diagnosis_4 diagnosis_5 diagnosis_6 diagnosis_7 diagnosis_8 diagnosis_9 diagnosis_10 diagnosis_11 diagnosis_12 diagnosis_13
1  123           1           3        <NA>        <NA>        <NA>        <NA>        <NA>        <NA>        <NA>         <NA>         <NA>         <NA>         <NA>
2 5345           2           3           1        <NA>        <NA>        <NA>        <NA>        <NA>        <NA>         <NA>         <NA>         <NA>         <NA>
3  234           3        <NA>        <NA>        <NA>        <NA>        <NA>        <NA>        <NA>        <NA>         <NA>         <NA>         <NA>         <NA>
4  453           4           1        <NA>        <NA>        <NA>        <NA>        <NA>        <NA>        <NA>         <NA>         <NA>         <NA>         <NA>
5 3656           5        <NA>        <NA>        <NA>        <NA>        <NA>        <NA>        <NA>        <NA>         <NA>         <NA>         <NA>         <NA>
6  345           1           4           3           1        <NA>        <NA>        <NA>        <NA>        <NA>         <NA>         <NA>         <NA>         <NA>
  age
1  54
2  65
3  23
4  22
5  33
6  77