aggregate with empty factor but keep row

问题

I had a similar questions with by() where I accepted the fact that I had to manually replace the resulting NAs. Now I would like to aggregate my data.frame and keep the structure. e.g. My larger data set has factors for 100 countries * 10 years * 5 segments, so it should reduce to 5000 rows. But sometimes some of the segment factors are empty and i only get <5000 rows. I cannot get my head around it...

My MWE still applies:

#All 3 categories are used
df1<-data.frame( val=rep(seq(1:4),3), factor=cut(rep(seq(1:4),3),breaks=c(1,2,3,4), include.lowest = TRUE, ordered_results=True , labels=LETTERS[1:3]))
# Thirds category is not used
df2<-data.frame( val=rep(seq(1:3),4), factor=cut(rep(seq(1:3),4),breaks=c(1,2,3,4), include.lowest = TRUE, ordered_results=True , labels=LETTERS[1:3]))

#df1 reduces to 3 rows as each category is used
aggregate(df1$val,list(df1$factor),sum)
#df2 reduces to 2 rows because C is empty
aggregate(df2$val,list(df2$factor),sum)
#I would like
data.frame(Group.1=LETTERS[1:3], x=c(12,12,0))

  Group.1  x
1       A 12
2       B 12
3       C  0

回答1:

# create dataset
df2 <- data.frame( val=rep(seq(1:3),4), factor=cut(rep(seq(1:3),4),breaks=c(1,2,3,4), include.lowest = TRUE, ordered_results=True , labels=LETTERS[1:3]))

library(dplyr)

levels(df2$factor) %>%                                    # get distinct levels of the factor variable
  data.frame(factor = .) %>%                              # create a data frame
  left_join(df2 %>%                                       # join with
            group_by(factor) %>%                             # for each value that exists
            summarise(x = sum(val)), by = "factor") %>%      # sum column val
  mutate(x = coalesce(x, 0L))                             # replace NAs with 0s

#   factor  x
# 1      A 12
# 2      B 12
# 3      C  0

Or without any package

dd = merge(data.frame(Group.1 = levels(df2$factor)), 
           aggregate(df2$val,list(df2$factor),sum), all.x = T)
dd$x = ifelse(is.na(dd$x), 0, dd$x)
dd

#   Group.1  x
# 1       A 12
# 2       B 12
# 3       C  0

Or using data.table package to check if it's faster

library(data.table)

# assuming you start with a data frame
df2 <- data.frame( val=rep(seq(1:3),4), factor=cut(rep(seq(1:3),4),breaks=c(1,2,3,4), include.lowest = TRUE, ordered_results=True , labels=LETTERS[1:3]))

# create a data table with all unique values of the variable "factor" and an index (key) on that variable
dt_levels = data.table(factor = levels(df2$factor), key = "factor")

# make df2 a data table with an index on column "factor" and aggregate
dt_sum = setDT(df2, key = "factor")[, list(Sum = sum(val)), by = "factor"]

# left join the two data tables and replace NA values with 0s
dt_result = dt_sum[dt_levels][, Sum := ifelse(is.na(Sum), 0, Sum)]

dt_result[]

#    factor Sum
# 1:      A  12
# 2:      B  12
# 3:      C   0

回答2:

You can use the complete function from tidyr to explicitly show the missing values in the results:

library(dplyr); library(tidyr)
df2 %>% 
        group_by(factor) %>% 
        summarise(x = sum(val)) %>% 
        complete(factor, fill = list(x = 0))

# Source: local data frame [3 x 2]

#   factor     x
#   <fctr> <dbl>
# 1      A    12
# 2      B    12
# 3      C     0

With aggregate function:

tidyr::complete(aggregate(df2$val,list(df2$factor),sum), Group.1, fill = list(x = 0))

# Source: local data frame [3 x 2]

#  Group.1     x
#   <fctr> <dbl>
#1       A    12
#2       B    12
#3       C     0

回答3:

So this is pretty basic, but I just made a new data frame with 2 columns. One with each factor level, and one with all 0. Then I used rbind to but my new frame and df2 together, and then used aggregate.

df2 <- data.frame( val=rep(seq(1:3),4), factor=cut(rep(seq(1:3),4),breaks=c(1,2,3,4), include.lowest = TRUE, ordered_results=True , labels=LETTERS[1:3]))

dat <- data.frame(val = 0, factor = levels(df2$factor))

df3 <- rbind(df2,dat)

aggregate(. ~ factor,df3,sum)

#  factor val
#1      A  12
#2      B  12
#3      C   0

来源：https://stackoverflow.com/questions/38898276/aggregate-with-empty-factor-but-keep-row

标签

aggregate