通过将值与列名匹配来填充 data.frame

问题描述

限时删除！！

我有两个 data.frames:

I have two data.frames:

df1 (空，但具有特定的姓氏)

df1 (empty, but with specific colnames)

apple orange banana pear grape
0      0       0       0     0

df2

fruit1  count1 fruit2 count2
apple   2      pear   1
grape  4      orange 2
banana 1      NA     NA

这是我想要的输出:

apples oranges bananas pears grapes
2      0       0       1     0
0      2       0       0     4
0      0       1       0     0

我考虑过采取以下措施:

I've considered doing something along the lines of:

for f (in range(nrow(df2))){
  for (i in range(ncol(df1))){
     if(fruit1==columnName[i]){
         df1[f,i]<-count1
         ect...

但是，我正在处理一个相当大的数据集，而这似乎并不是正确的方法.

However, i'm dealing with a rather large dataset, and this doesn't seem like the right way to do it.

推荐答案

data.table选项:

A data.table option:

library(data.table)
setDT(df2)

# add row number
df2[, r := .I]

# "melt" common columns together
cols = c("fruit", "count")
m2 = melt(df2, measure=patterns(cols), value.name=cols)

# add unobserved fruits, if any
m2[, fruit := factor(fruit, levels = names(df1))]

# "cast" each fruit to its own column, ignoring NA rows
dcast(m2[!is.na(fruit)], r ~ fruit, fill = 0L, drop = FALSE)


   r apple orange banana pear grape
1: 1     2      0      0    1     0
2: 2     0      2      0    0     4
3: 3     0      0      1    0     0

factor 只是为了防止您在 df1 中有一些额外的级别而不会出现在 df2 中.如果一行中的 fruit1 和 fruit2 均为空白，则必须弄清楚如何扩展此方法.

The factor thing is just in case you have some extra levels in df1 that don't appear in df2. If you have a row where both fruit1 and fruit2 are blank, you'll have to figure out how you want to extend this approach.

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