r：汇总group_by之后的rowSums

Question

I've tried searching a number of posts on SO but I'm not sure what I'm doing wrong here, and I imagine the solution is quite simple. I'm trying to group a dataframe by one variable and figure the mean of several variables within that group.

Here is what I am trying:

head(airquality)
target_vars = c("Ozone","Temp","Solar.R")
airquality %>% group_by(Month) %>% select(target_vars) %>% summarise(rowSums(.))

But I get the error that my lenghts don't match. I've tried variations using mutate to create the column or summarise_all , but neither of these seem to work. I need the row sums within group, and then to compute the mean within group (yes, it's nonsensical here).

Also, I want to use select because I'm trying to do this over just certain variables.

I'm sure this could be a duplicate, but I can't find the right one.

EDIT FOR CLARITY Sorry, my original question was not clear. Imagine the grouping variable is the calendar month, and we have v1 , v2 , and v3 . I'd like to know, within month , what was the average of the sums of v1 , v2 , and v3 . So if we have 12 months, the result would be a 12x1 dataframe. Here is an example if we just had 1 month:

Month v1 v2 v3 Sum 
1      1  1  0   2
1      1  1  1   3
1      1  0  0   3

Then the result would be:

Month  Average
1           8/3

Answer 1

This seems to deliver what you want. It's regular R. The sapply function keeps the months separated by "name". The sum function applied to each dataframe will not keep the column sums separate. (Correction # 2: used only target_vars):

sapply( split( airquality[target_vars], airquality$Month), sum, na.rm=TRUE)
    5     6     7     8     9 
 7541  8343 10849  8974  8242 

If you wanted the per number of variable results, then you would divide by the number of variables:

sapply( split( airquality[target_vars], airquality$Month), sum, na.rm=TRUE)/
                                                           (length(target_vars))
       5        6        7        8        9 
2513.667 2781.000 3616.333 2991.333 2747.333 

Answer 2

You can try:

library(tidyverse)
airquality %>% 
  select(Month, target_vars) %>% 
  gather(key, value, -Month) %>% 
  group_by(Month) %>%
  summarise(n=length(unique(key)),
            Sum=sum(value, na.rm = T)) %>% 
  mutate(Average=Sum/n)
# A tibble: 5 x 4
  Month     n   Sum  Average
  <int> <int> <int>    <dbl>
1     5     3  7541 2513.667
2     6     3  8343 2781.000
3     7     3 10849 3616.333
4     8     3  8974 2991.333
5     9     3  8242 2747.333

The idea is to convert the data from wide to long using tidyr::gather() , then group by Month and calculate the sum and the average.

Answer 3

Perhaps this is what you're looking for

library(dplyr)
library(purrr)
library(tidyr)   # forgot this in original post
airquality %>%
  group_by(Month) %>% 
  nest(Ozone, Temp, Solar.R, .key=newcol) %>%
  mutate(newcol = map_dbl(newcol, ~mean(rowSums(.x, na.rm=TRUE))))

# A tibble: 5 x 2
  # Month   newcol
  # <int>    <dbl>
# 1     5 243.2581
# 2     6 278.1000
# 3     7 349.9677
# 4     8 289.4839
# 5     9 274.7333  

I've never encountered a situation where all the answers disagreed. Here's some validation (at least I think) for the 5th month

airquality %>%
  filter(Month == 5) %>%
  select(Ozone, Temp, Solar.R) %>%
  mutate(newcol = rowSums(., na.rm=TRUE)) %>%
  summarise(sum5 = sum(newcol), mean5 = mean(newcol))

#   sum5    mean5
# 1 7541 243.2581