Dplyr Mutate_each用于配对列 [英] Dplyr Mutate_each for paired sets of columns
问题描述
有没有办法使用dplyr :: mutate_each来实现以下转换？
Is there a way to achieve the following transformation using dplyr::mutate_each?
data.frame(x1 = 1:5, x2 = 6:10, y1 = rnorm(5), y2 = rnorm(5)) %>%
mutate(diff1 = x1 - y1, diff2 = x2 - y2)
## x1 x2 y1 y2 diff1 diff2
## 1 1 6 1.03645018 -0.8602099 -0.03645018 6.860210
## 2 2 7 -1.10790835 1.6912875 3.10790835 5.308712
## 3 3 8 0.95452119 2.7232657 2.04547881 5.276734
## 4 4 9 0.01370762 1.6385765 3.98629238 7.361424
## 5 5 10 0.19354354 -1.0464360 4.80645646 11.046436
我知道这是一个微不足道的例子，很容易完成，如我所描述的，但我正在尝试用更大的列来完成类似的事情。
I realize this is a trivial example and is easily done as I've described, but I'm trying to accomplish a similar thing with a much larger set of columns.
谢谢
推荐答案
根据@Gregor在评论中的提及，如果你想工作使用 dplyr ，最好获取您的数据在一个整洁的格式。这是一个想法：
As per mentionned by @Gregor in the comments, if you want to work with dplyr, it would be better to get your data in a tidy format. Here's an idea:
library(dplyr)
library(tidyr)
df %>%
add_rownames() %>%
gather(key, val, -rowname) %>%
separate(key, c("var", "num"), "(?<=[a-z]) ?(?=[0-9])") %>%
spread(var, val) %>%
mutate(diff = x - y)
其中：
#Source: local data frame [10 x 5]
#
# rowname num x y diff
# (chr) (chr) (dbl) (dbl) (dbl)
#1 1 1 1 1.03645018 -0.03645018
#2 1 2 6 -0.86020990 6.86020990
#3 2 1 2 -1.10790835 3.10790835
#4 2 2 7 1.69128750 5.30871250
#5 3 1 3 0.95452119 2.04547881
#6 3 2 8 2.72326570 5.27673430
#7 4 1 4 0.01370762 3.98629238
#8 4 2 9 1.63857650 7.36142350
#9 5 1 5 0.19354354 4.80645646
#10 5 2 10 -1.04643600 11.04643600
如果由于某种原因，您仍然希望在执行操作后使用宽格式的数据，则可以添加到管道中：
If for some reason you still want the data in wide format after performing the operation, you could add to the pipe:
gather(key, value, -(rowname:num)) %>%
unite(key_num, key, num, sep = "") %>%
spread(key_num, value)
哪个会给：
#Source: local data frame [5 x 7]
#
# rowname diff1 diff2 x1 x2 y1 y2
# (chr) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl)
#1 1 -0.03645018 6.860210 1 6 1.03645018 -0.8602099
#2 2 3.10790835 5.308713 2 7 -1.10790835 1.6912875
#3 3 2.04547881 5.276734 3 8 0.95452119 2.7232657
#4 4 3.98629238 7.361423 4 9 0.01370762 1.6385765
#5 5 4.80645646 11.046436 5 10 0.19354354 -1.0464360
数据 / p>
Data
df <- structure(list(x1 = 1:5, x2 = 6:10, y1 = c(1.03645018, -1.10790835,
0.95452119, 0.01370762, 0.19354354), y2 = c(-0.8602099, 1.6912875,
2.7232657, 1.6385765, -1.046436)), .Names = c("x1", "x2", "y1",
"y2"), class = "data.frame", row.names = c("1", "2", "3", "4", "5"))
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