R-用任何其他列中的值填充列 [英] R - Fill Column with values from any other columns

问题描述

我有一个5列的数据框:4列具有值，而1列为空.我想用4个列中任何一个的任何值填充空列.

I have a data frame of 5 columns: 4 columns have values and the 1 column that is empty. I want to fill the empty column with any value from any of the 4 columns.

假设这是我的数据框df:

Col1 Col2 Col3 Col4 Col5
  11   11    
   2         2    2
       23
   4         4
       15        15

我希望我的结果看起来像这样:

I want my result to look like this:

Col1 Col2 Col3 Col4 Col5
  11   11             11
   2         2    2    2
       23             23
   4         4         4
       15        15   15

编辑，我应用了每个人提供的答案，但是由于某些原因，它仍然无法正常工作.如果有帮助，这是我的实际数据中的dput(head(df)):

EDIT I applied the answers provided by everyone, but it still isn't working for some reason. If it helps, this is the dput(head(df)) of my actual data:

structure(list(Treat_One = c("      ", "5 2012", "4 2008", "4 2010", 
"      ", "2 2008"), Treat_Two = c("8 2010", "5 2012", "4 2008", 
"4 2010", "8 2011", "2 2008"), Treat_Three = c("      ", "5 2012", 
"4 2008", "4 2010", "8 2011", "2 2008"), Treat_Four = c("      ", 
"      ", "      ", "      ", "      ", "      ")), .Names = c("Treat_One", 
"Treat_Two", "Treat_Three", "Treat_Four"), row.names = c(NA, 
6L), class = "data.frame")

编辑包含的str(df)

'data.frame':   209 obs. of  4 variables:
 $ Treat_One  : chr  "      " "5 2012" "4 2008" "4 2010" ...
 $ Treat_Two  : chr  "8 2010" "5 2012" "4 2008" "4 2010" ...
 $ Treat_Three: chr  "      " "5 2012" "4 2008" "4 2010" ...
 $ Treat_Four : chr  "      " "      " "      " "      " ...

推荐答案

基于OP提供的新数据，我们可以使用trimws

Based on the new data provided by the OP we can remove leading/trailing white spaces with trimws

df$Treat_Four <- apply(df, 1, function(x) sample(x[trimws(x) != ""], 1))
df

#    Treat_One Treat_Two Treat_Three Treat_Four
#1              8 2010                 8 2010
#2    5 2012    5 2012      5 2012     5 2012
#3    4 2008    4 2008      4 2008     4 2008
#4    4 2010    4 2010      4 2010     4 2010
#5              8 2011      8 2011     8 2011
#6    2 2008    2 2008      2 2008     2 2008

原始答案

我们可以逐行使用apply并取不等于空字符串的元素中的1 sample

We can use apply row-wise and take 1 sample of the element which is not equal to empty string

df$Col5 <- apply(df, 1, function(x) sample(x[x != ""], 1))
df
#  Col1 Col2 Col3 Col4 Col5
#1    1    1              1
#2    2         2    2    2
#3         3              3
#4    4         4         4
#5         5         5    5

---

如果有NA个值而不是空格，我们可以使用相同的逻辑

---

If there are NA values and not blanks we can use the same logic

apply(df, 1, function(x) sample(x[!is.na(x)], 1))

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