聚合为空因子但保留行 [英] aggregate with empty factor but keep row

问题描述

我对by（）有类似的问题，我接受了必须手动替换生成的NA的事实。现在，我想聚合我的data.frame并保留结构。例如我的较大数据集包含100个国家/地区* 10年* 5个细分的因素，因此应减少到5000行。但是有时某些细分市场因素是空的，我只能得到<5000行。我没办法解决...

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...

我的MWE仍然适用：

#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

推荐答案

# 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

或者不带任何包装

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

或使用 data.table 包检查其速度是否更快

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         

这篇关于聚合为空因子但保留行的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！