如何根据查找表替换列中的字符串值 [英] How to replace string values in a column based on a lookup table

问题描述

我有这个df.我想根据查找表替换"1.1"，"2.1"，"3.2"等的每个实例.

I have this df. I'd like to replace every instance of "1.1", "2.1" "3.2", etc. based on a lookup table.

查询表

所需的输出

我查看了替换数据框中的值基于查找表不能解决我的问题. 预先感谢！

I looked at Replace values in a dataframe based on lookup table it doesn't solve my problem. Thanks in advance!

推荐答案

选项为gsubfn.我们将一个或多个数字与一个点([0-9.]+)匹配为模式，并在替换中传递从第二个数据集('df2')创建的键/值对的list.对于带有键"的匹配模式，它将替换字符串中相应的值"

An option would be gsubfn. We match one or more digits with a dot ([0-9.]+) as the pattern and in the replacement, pass a list of key/value pairs created from the second dataset ('df2'). For the matching pattern with the 'keys', it replace the corresponding 'value' within the string

library(gsubfn)
df1$Node <- gsubfn("([0-9.]+)", as.list(setNames(df2$Label,df2$Node)), df1$Node)
df1$Node
#[1] "One one > Two one > Three two" "One two > Two two > Three one" "One one > Two two > Three one" "One two > Two one > Three two"
#[5] "One one > Two two > Three two"

数据

df1 <- data.frame(ID = 1:5, Node = c("1.1 > 2.1 > 3.2", "1.2 > 2.2 > 3.1", "1.1 > 2.2 > 3.1", "1.2 > 2.1 > 3.2", "1.1 > 2.2 > 3.2"), stringsAsFactors = FALSE)

df2 <- data.frame(Label =  c("One one", "One two", "Two one", "Two two", "Three one", "Three two"), Node = c("1.1", "1.2", "2.1", "2.2", "3.1", "3.2"), stringsAsFactors = FALSE)

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