替换数据帧中的所有特定值

Having a data frame, how do I go about replacing all particular values along all rows and columns. Say for example I want to replace all empty records with NA's (without typing the positions):

有了数据框架，我如何替换所有行和列上的所有特定值。例如，我想用NA替换所有空记录(不输入位置):

df <- data.frame(list(A=c("", "xyz", "jkl"), B=c(12, "", 100)))    A   B1      122  xyz    3  jkl 100

Expected result:

预期结果:

    A   B1  NA   122  xyz  NA  3  jkl  100

4 个解决方案

#1

Like this:

是这样的:

> df[df==""]<-NA> df     A    B1 <NA>   122  xyz <NA>3  jkl  100

#2

Since PikkuKatja and glallen asked for a more general solution and I cannot comment yet, I'll write an answer. You can combine statements as in:

由于pikkkukatja和glallen要求一个更一般的解决方案，我现在还不能评论，我将写一个答案。您可以将语句合并为:

> df[df=="" | df==12] <- NA> df     A    B1  <NA> <NA>2  xyz  <NA>3  jkl  100

For factors, zxzak's code already yields factors:

对于因子，zxzak的代码已经产生因子:

> df <- data.frame(list(A=c("","xyz","jkl"), B=c(12,"",100)))> str(df)'data.frame':   3 obs. of  2 variables: $ A: Factor w/ 3 levels "","jkl","xyz": 1 3 2 $ B: Factor w/ 3 levels "","100","12": 3 1 2

If in trouble, I'd suggest to temporarily drop the factors.

如果遇到麻烦，我建议暂时放弃这些因素。

df[] <- lapply(df, as.character)

#3

We can use data.table to get it quickly.First create df without factors,

我们可以使用的数据。桌子可以让它快速移动。首先创建无因子df，

df <- data.frame(list(A=c("","xyz","jkl"), B=c(12,"",100)), stringsAsFactors=F)

Now you can use

现在你可以使用

setDT(df)for (jj in 1:ncol(df)) set(df, i = which(df[[jj]]==""), j = jj, v = NA)

and you can convert it back to a data.frame

你可以把它转换回data.frame

setDF(df)

If you only want to use data.frame and keep factors it's more difficult, you need to work with

如果你只想用data.frame且保存因子会更困难，那你就需要使用它

levels(df$value)[levels(df$value)==""] <- NA

where value is the name of every column. You need to insert it in a loop.

其中value是每个列的名称。您需要将它插入到循环中。

#4

If you want to replace multiple values in a data frame, looping through all columns might help.

如果您想要在数据帧中替换多个值，那么遍历所有列可能会有所帮助。

Say you want to replace "" and 100:

假设你想要替换“”和100:

na_codes <- c(100, "")for (i in seq_along(df)) {    df[[i]][df[[i]] %in% na_codes] <- NA}

#1