5

I wish to obtain weighted means by group for each of several (actually about 60) columns. This question is very similar to: repeatedly applying ave for computing group means in a data frame just asked.

I have come up with two ways to obtain the weighted means so far:

  1. use a separate sapply statement for each column
  2. place an sapply statement inside a for-loop

However, I feel there must be a way to insert an apply statement inside the sapply statement or vice versa, thereby eliminating the for-loop. I have tried numerous permutations without success. I also looked at the sweep function.

Here is the code I have so far.

df <- read.table(text= "
          region    state  county  weights y1980  y1990  y2000
             1        1       1       10     100    200     50
             1        1       2        5      50    100    200
             1        1       3      120    1000    500    250
             1        1       4        2      25    100    400
             1        1       4       15     125    150    200

             2        2       1        1      10     50    150
             2        2       2       10      10     10    200
             2        2       2       40      40    100     30
             2        2       3       20     100    100     10
", header=TRUE, na.strings=NA)

# add a group variable to the data set

group <- paste(df$region, '_', df$state, '_', df$county, sep = "")
df    <- data.frame(group, df)

# obtain weighted averages for y1980, y1990 and y2000 
# one column at a time using one sapply per column

sapply(split(df, df$group), function(x) weighted.mean(x$y1980, w = x$weights))
sapply(split(df, df$group), function(x) weighted.mean(x$y1990, w = x$weights))
sapply(split(df, df$group), function(x) weighted.mean(x$y2000, w = x$weights))

# obtain weighted average for y1980, y1990 and y2000
# one column at a time using a for-loop

y <- matrix(NA, nrow=7, ncol=3)
group.b <- df[!duplicated(df$group), 1]

for(i in 6:8) { 

    y[,(i-5)] <- sapply(split(df[,c(1:5,i)], df$group), function(x) weighted.mean(x[,6], w = x$weights))

}

# add weighted averages to the original data set

y2 <- data.frame(group.b, y)
colnames(y2) <- c('group','ave1980','ave1990','ave2000')
y2

y3 <- merge(df, y2, by=c('group'), all = TRUE)
y3

Sorry for all of my questions lately, and thank you for any advice.

EDITED to show y3

  group region state county weights y1980 y1990 y2000   ave1980  ave1990  ave2000
1 1_1_1      1     1      1      10   100   200    50  100.0000 200.0000  50.0000
2 1_1_2      1     1      2       5    50   100   200   50.0000 100.0000 200.0000
3 1_1_3      1     1      3     120  1000   500   250 1000.0000 500.0000 250.0000
4 1_1_4      1     1      4       2    25   100   400  113.2353 144.1176 223.5294
5 1_1_4      1     1      4      15   125   150   200  113.2353 144.1176 223.5294
6 2_2_1      2     2      1       1    10    50   150   10.0000  50.0000 150.0000
7 2_2_2      2     2      2      10    10    10   200   34.0000  82.0000  64.0000
8 2_2_2      2     2      2      40    40   100    30   34.0000  82.0000  64.0000
9 2_2_3      2     2      3      20   100   100    10  100.0000 100.0000  10.0000
Community
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Mark Miller
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  • Keep the questions coming, but I (we?) would appreciate it, if you provided some dummy data and demonstrate how the desired result would look like. – Roman Luštrik Jan 03 '13 at 20:03
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    Sorry. I have added the contents of the data set 'y3' to my post. The dummy data are in the data set 'df' and the code seems to run on my computer. – Mark Miller Jan 03 '13 at 20:08

2 Answers2

6

I suggest to use package data.table:

library(data.table)
dt <- as.data.table(df)
dt2 <- dt[,lapply(.SD,weighted.mean,w=weights),by=list(region,state,county)]
print(dt2)

   region state county   weights     y1980    y1990    y2000
1:      1     1      1  10.00000  100.0000 200.0000  50.0000
2:      1     1      2   5.00000   50.0000 100.0000 200.0000
3:      1     1      3 120.00000 1000.0000 500.0000 250.0000
4:      1     1      4  13.47059  113.2353 144.1176 223.5294
5:      2     2      1   1.00000   10.0000  50.0000 150.0000
6:      2     2      2  34.00000   34.0000  82.0000  64.0000
7:      2     2      3  20.00000  100.0000 100.0000  10.0000

If you want you can merge with the original data.table afterwards:

merge(dt,dt2,by=c("region","state","county"))

   region state county weights.x y1980.x y1990.x y2000.x weights.y   y1980.y  y1990.y  y2000.y
1:      1     1      1        10     100     200      50  10.00000  100.0000 200.0000  50.0000
2:      1     1      2         5      50     100     200   5.00000   50.0000 100.0000 200.0000
3:      1     1      3       120    1000     500     250 120.00000 1000.0000 500.0000 250.0000
4:      1     1      4         2      25     100     400  13.47059  113.2353 144.1176 223.5294
5:      1     1      4        15     125     150     200  13.47059  113.2353 144.1176 223.5294
6:      2     2      1         1      10      50     150   1.00000   10.0000  50.0000 150.0000
7:      2     2      2        10      10      10     200  34.00000   34.0000  82.0000  64.0000
8:      2     2      2        40      40     100      30  34.00000   34.0000  82.0000  64.0000
9:      2     2      3        20     100     100      10  20.00000  100.0000 100.0000  10.0000
Roland
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1

I figured out how to nest sapply inside apply to obtain weighted averages by group and column without using an explicit for-loop. Below I provide the data set, the apply statement and an explanation of how the apply statement works.

Here is the data set from the original post:

df <- read.table(text= "
          region    state  county  weights y1980  y1990  y2000
             1        1       1       10     100    200     50
             1        1       2        5      50    100    200
             1        1       3      120    1000    500    250
             1        1       4        2      25    100    400
             1        1       4       15     125    150    200

             2        2       1        1      10     50    150
             2        2       2       10      10     10    200
             2        2       2       40      40    100     30
             2        2       3       20     100    100     10
", header=TRUE, na.strings=NA)

# add a group variable to the data set

group <- paste(df$region, '_', df$state, '_', df$county, sep = "")
df    <- data.frame(group, df)

Here is the apply / sapply code to obtain the desired weighted means.

apply(df[,6:ncol(df)], 2, function(x) {sapply(split(data.frame(df[,1:5], x), df$group), function(y) weighted.mean(y[,6], w = y$weights))})

Here is an explanation of the above apply / sapply statement:

  1. Note that the apply statement selects columns 6 through 8 of df one at a time.

  2. For each of those three columns I create a new data frame combining that individual column with the first five columns of df.

  3. Then I split each of those new 6-column data frames into chunks by the grouping variable df$group.

  4. Once a data frame of six columns has been split into its individual chunks I calculate the weighted mean for the last column (the 6th column) of each chunk.

Here is the result:

          y1980    y1990    y2000
1_1_1  100.0000 200.0000  50.0000
1_1_2   50.0000 100.0000 200.0000
1_1_3 1000.0000 500.0000 250.0000
1_1_4  113.2353 144.1176 223.5294
2_2_1   10.0000  50.0000 150.0000
2_2_2   34.0000  82.0000  64.0000
2_2_3  100.0000 100.0000  10.0000

Using package data.table is nice, but until I become more familiar with its syntax and how that syntax differs from the syntax of data.frame I thought it would be good to know how to use apply and sapply to do the same thing. Now I can use both approaches, plus the approaches in the original post, to check one against the others and learn more about all of them.

Mark Miller
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  • This solution is a bit convoluted and its performance is probably not very good (but I haven't benchmarked). You are basically trying to reinvent the wheel. If you want nice syntax you should look into package `plyr`. I demonstrated the `data.table` solution because it's much faster with big datasets. All split-apply-combine functions allow you to specify several grouping variables. – Roland Jan 05 '13 at 14:00
  • I just benchmarked both solutions. Yours was about 2.5 times faster. Thank you for the suggestions. – Mark Miller Jan 05 '13 at 15:57