有没有办法使用dplyr基于另一列的group_by除以创建新列？
[英]Is there a way using dplyr to create a new column based on dividing by group_by of another column?
问题描述
I am trying to create a new column by dividing a column A of integers (data1/2/3 below) by the mode of column A when grouped by another column B of integers (group1/2 below) 
我试图通过将整数列A（下面的数据1/2/3）除以整列的另一列B（下面的组1/2）分组列A的模式来创建一个新列
group1=rep(1:5,each=2)
group2=rep(6:10, each=2)
data1=c(1,1,1,1,1,4,5,6,3,8)
data2=c(5,4,5,7,8,5,2,1,1,5)
data3=c(6,6,8,9,5,4,3,3,1,1)
DF=data.frame(group1,group2,data1,data2,data3)
group1 group2 data1 data2 data3
1 1 6 1 5 6
2 1 6 1 4 6
3 2 7 1 5 8
4 2 7 1 7 9
5 3 8 1 8 5
6 3 8 4 5 4
7 4 9 5 2 3
8 4 9 6 1 3
9 5 10 3 1 1
10 5 10 8 5 1
I have been successful in doing this one column at a time (see code below), but I would like to be able to generalize it: 我一次成功完成了这一列（见下面的代码），但我希望能够概括它：
DF %>%
group_by(group2) %>%
mutate(group2_mode = as.integer(head(names(sort(table(data2))),1))) %>%
mutate(group2_data2 = data2/group2_mode) %>%
#select(-c(group1_mode)) %>%
ungroup()
# A tibble: 10 x 7
group1 group2 data1 data2 data3 group2_mode group2_data2
<int> <int> <dbl> <dbl> <dbl> <int> <dbl>
1 1 6 1 5 6 4 1.25
2 1 6 1 4 6 4 1
3 2 7 1 5 8 5 1
4 2 7 1 7 9 5 1.4
5 3 8 1 8 5 5 1.6
6 3 8 4 5 4 5 1
7 4 9 5 2 3 1 2
8 4 9 6 1 3 1 1
9 5 10 3 1 1 1 1
10 5 10 8 5 1 1 5
This works but is clunky when written out for each data/group combination. 这可行但在为每个数据/组组合写出时很笨拙。
I have tried iterating through for loops as follows: 我已经尝试迭代for循环，如下所示：
for (i in colnames(DF[,3:5])){
for (k in colnames(DF[,1:2])){
DF %>%
group_by(k) %>%
mutate(paste(c(k,"_",i), collapse = '') <- i/as.integer(head(names(sort(table(i))),1)))
}
}
And receive the following error: 并收到以下错误：
Error: Column `k` is unknown
I expect the output to be similar to the first code chunk above but for each data/group combination. 我希望输出类似于上面的第一个代码块，但是对于每个数据/组组合。 I have also tried labeling all of the mutated columns in the for loop the same thing, but that also results in the same error. 我也尝试在for循环中标记所有变异列同样的东西，但这也会导致相同的错误。 I suspect the issue lies in the group_by statement, but I can't figure out how. 我怀疑问题出在group_by语句中，但我无法弄清楚如何。
Thank you for your time 感谢您的时间
3 个解决方案
解决方案1
2 2019-08-29 00:04:44
Borrowing from here , we can define a helper mode function: 借用这里 ，我们可以定义一个辅助mode函数：
mode <- function(codes){
which.max(tabulate(codes))
}
Then: 然后：
DF %>%
group_by(group2) %>%
mutate_at(vars(matches("data")), ~. / mode(.))
[This should work, in theory, but this mode function seems to work differently than yours, and I don't see how to resolve yet.] [理论上这应该有效，但这种模式功能似乎与你的不同，我看不出如何解决。]
Edit: To do this with a few multiple groups, you could create new columns like so: 编辑：要使用几个多个组执行此操作，您可以创建新列，如下所示：
DF %>%
group_by(group1) %>%
mutate_at(vars(matches("data")),
.funs = list(gp1 = ~. / mode(.))) %>%
group_by(group2) %>%
mutate_at(vars(matches("data")),
.funs = list(gp2 = ~. / mode(.)))
# A tibble: 10 x 14
# Groups: group2 [5]
group1 group2 data1 data2 data3 data1_gp1 data2_gp1 data3_gp1 data1_gp2 data2_gp2 data3_gp2 data1_gp1_gp2 data2_gp1_gp2 data3_gp1_gp2
<int> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 6 1 5 6 1 1.25 1 1 1.25 1 1 1.25 1
2 1 6 1 4 6 1 1 1 1 1 1 1 1 1
3 2 7 1 5 8 1 1 1 1 1 1 1 1 1
4 2 7 1 7 9 1 1.4 1.12 1 1.4 1.12 1 1.4 1.12
5 3 8 1 8 5 1 1.6 1.25 1 1.6 1.25 1 1.6 1.25
6 3 8 4 5 4 4 1 1 4 1 1 4 1 1
7 4 9 5 2 3 1 2 1 1 2 1 1 2 1
8 4 9 6 1 3 1.2 1 1 1.2 1 1 1.2 1 1
9 5 10 3 1 1 1 1 1 1 1 1 1 1 1
10 5 10 8 5 1 2.67 5 1 2.67 5 1 2.67 5 1
If you have many groups, then we might want to create a function for this. 如果你有很多组，那么我们可能想为此创建一个函数。 This one mostly works, except for the naming step -- I want my group selection to also provide the name for the new column labels. 除了命名步骤之外，这个主要有效 - 我希望我的组选择也提供新列标签的名称。 := didn't seem to work for me here, which seems otherwise to be the way to name new columns in tidyeval. :=似乎在这里对我不起作用，这似乎是在tidyeval中命名新列的方法。 Can someone help me here? 有人可以帮我吗？
add_grouped_medians <- function(df, group) {
suffix = !!group # This part seems to be missing the right
# syntax. I want to make the group input available to the
# .funs list below....
df %>%
group_by(!! group) %>%
mutate_at(vars(matches("data")),
.funs = list( suffix = ~. / mode(.)))
}
Note how the output uses "suffix" literally instead of using the group name in its place: 请注意输出如何使用“suffix”字面而不是在其位置使用组名：
> DF %>% add_grouped_medians(group1, "gp1")
# A tibble: 10 x 9
# Groups: <int> [5]
group1 group2 data1 data2 data3 `<int>` data1_suffix data2_suffix data3_suffix
<int> <int> <dbl> <dbl> <dbl> <int> <dbl> <dbl> <dbl>
1 1 6 1 5 6 1 1 1.25 1
2 1 6 1 4 6 1 1 1 1
3 2 7 1 5 8 2 1 1 1
4 2 7 1 7 9 2 1 1.4 1.12
5 3 8 1 8 5 3 1 1.6 1.25
6 3 8 4 5 4 3 4 1 1
7 4 9 5 2 3 4 1 2 1
8 4 9 6 1 3 4 1.2 1 1
9 5 10 3 1 1 5 1 1 1
10 5 10 8 5 1 5 2.67 5 1
解决方案2
1 2019-08-29 01:45:50
You could try some tidy evaluation. 你可以尝试一些整洁的评估。 The definition of Mode is taken from here . Mode的定义取自此处 。
Mode <- function(x) {
ux <- unique(x)
ux[which.max(tabulate(match(x, ux)))]
}
We can use grep to separate group and data columns. 我们可以使用grep来分隔group和data列。 Then use a for loop over them 然后在它们上面使用for循环
library(dplyr)
library(rlang)
group_cols <- grep("^group", names(DF), value = TRUE)
data_cols <- grep("^data", names(DF), value = TRUE)
for (col in seq_along(group_cols)) {
data <- sym(data_cols[col])
DF <- DF %>%
group_by_at(group_cols[col]) %>%
mutate(!!paste0("group", col, "mode") := !!data/Mode(!!data))
}
DF
# group1 group2 data1 data2 data3 group1mode group2mode
# <int> <int> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 1 6 1 5 6 1 1
# 2 1 6 1 4 6 1 0.8
# 3 2 7 1 5 8 1 1
# 4 2 7 1 7 9 1 1.4
# 5 3 8 1 8 5 1 1
# 6 3 8 4 5 4 4 0.625
# 7 4 9 5 2 3 1 1
# 8 4 9 6 1 3 1.2 0.5
# 9 5 10 3 1 1 1 1
#10 5 10 8 5 1 2.67 5
Few things to note, as already mentioned by @Jon Spring your Mode calculation is different than the standard one. 很少有事情要注意，正如@Jon Spring已经提到的，你的模式计算与标准计算不同。 If needed you can change the above Mode to your way of calculating it. 如果需要，您可以将上述Mode更改为您的计算方式。 Also in reality I hope you would have same number of group and data columns (here they are unequal). 实际上我希望你有相同数量的group和data列（这里它们是不相等的）。
解决方案3
1 已采纳 2019-08-29 21:43:20
A base solution might be just as useful - I used the mode function suggested by @Jon Spring. 基本解决方案可能同样有用 - 我使用了@Jon Spring建议的mode功能。
mode <- function(codes){
which.max(tabulate(codes))
}
groups <- c('group1', 'group2')
datas <- c('data1', 'data2', 'data3')
for (grp in groups) {
for (col in datas) {
DF[, paste(col, grp, sep = '_')] <- ave(x = DF[[col]], DF[[grp]], FUN = function(x) x / mode(x))
}
}
group1 group2 data1 data2 data3 data1_group1 data2_group1 data3_group1 data1_group2 data2_group2 data3_group2
1 1 6 1 5 6 1.000000 1.25 1.000 1.000000 1.25 1.000
2 1 6 1 4 6 1.000000 1.00 1.000 1.000000 1.00 1.000
3 2 7 1 5 8 1.000000 1.00 1.000 1.000000 1.00 1.000
4 2 7 1 7 9 1.000000 1.40 1.125 1.000000 1.40 1.125
5 3 8 1 8 5 1.000000 1.60 1.250 1.000000 1.60 1.250
6 3 8 4 5 4 4.000000 1.00 1.000 4.000000 1.00 1.000
7 4 9 5 2 3 1.000000 2.00 1.000 1.000000 2.00 1.000
8 4 9 6 1 3 1.200000 1.00 1.000 1.200000 1.00 1.000
9 5 10 3 1 1 1.000000 1.00 1.000 1.000000 1.00 1.000
10 5 10 8 5 1 2.666667 5.00 1.000 2.666667 5.00 1.000
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