我有一张这样的表:
df <- read.table(text =
" Day city gender week
'day1' 'city1' 'M' 'one'
'day2' 'city2' 'M' 'two'
'day1' 'city3' 'F' 'two'
'day2' 'city4' 'F' 'two'",
header = TRUE, stringsAsFactors = FALSE)
我计算这样的汇总表:
daily_table <- setDT(df)[, .(Daily_Freq = .N,
men = sum(gender == 'M'),
women = sum(gender == 'F'),
city1 = sum(city == 'city1'),
city2 = sum(city == 'city2'),
city3 = sum(city == 'city3'),
city4 = sum(city == 'city4'),
city5 = sum(city == 'city5'))
, by = .(week,Day)]
制作此表:
week Day Daily_Freq men women city1 city2 city3 city4 city5
one day1 1 1 0 1 0 0 0 0
two day2 2 1 1 0 1 0 1 0
two day1 1 0 1 0 0 1 0 0
但是因为我有几个城市,所以我想使用带有名字的矢量:
cities <- c("city1","city2","city3","city4","city5")
请注意,我的向量中有5个城市,即使其中一个城市的出现次数为零,我希望它出现在我的最终表格中。 我该怎么办?
答案 0 :(得分:1)
为了确保R显示city5
,即使没有对该值的观察,请将其添加为因子级别:
setDT(df)
df[, city := factor(city,
levels = c("city1","city2","city3","city4","city5"))]
为了避免为city
的每个级别编写测试,您可以迭代city
级别,如下所示:
daily_table <- df[, c(.(Daily_Freq = .N,
men = sum(gender == 'M'),
women = sum(gender == 'F')),
lapply(setNames(levels(city), levels(city)),
function(x) sum(city == x))),
by = .(week,Day)]
daily_table
## week Day Daily_Freq men women city1 city2 city3 city4 city5
## 1: one day1 1 1 0 1 0 0 0 0
## 2: two day2 2 1 1 0 1 0 1 0
## 3: two day1 1 0 1 0 0 1 0 0