Question

我有一个看起来像这样的data.table：

> dt <- data.table(
  group1 = c("a", "a", "a", "b", "b", "b", "b"),
  group2 = c("x", "x", "y", "y", "z", "z", "z"),
  data1 = c(NA, rep(T, 3), rep(F, 2), "sometimes"),
  data2 = c("sometimes", rep(F,3), rep(T,2), NA))

> dt

   group1 group2     data1     data2
1:      a      x        NA sometimes
2:      a      x      TRUE     FALSE
3:      a      y      TRUE     FALSE
4:      b      y      TRUE     FALSE
5:      b      z     FALSE      TRUE
6:      b      z     FALSE      TRUE
7:      b      z sometimes        NA

我的目标是找到每个数据列中的非NA记录数，按group1和group2分组。

   group1 group2     data1     data2
1:      a      x         1         2
3:      a      y         1         1
4:      b      y         1         1
5:      b      z         3         2

我在处理数据集的另一部分时遗留了这些代码，该部分没有NA并且是合乎逻辑的：

dt[
  ,
  lapply(.SD, sum),
  by = list(group1, group2),
  .SDcols = c("data3", "data4")
]

但它不能使用NA值或非逻辑值。

Answer 1

dt[, lapply(.SD, function(x) sum(!is.na(x))), by = .(group1, group2)]
#   group1 group2 data1 data2
#1:      a      x     1     2
#2:      a      y     1     1
#3:      b      y     1     1
#4:      b      z     3     2

Answer 2

另一个替代方法是melt / dcast，以避免列操作。这将删除NAs并默认使用length功能

dcast(melt(dt, id = c("group1", "group2"), na.rm = TRUE), group1 + group2 ~ variable) 
# Aggregate function missing, defaulting to 'length'
#    group1 group2 data1 data2
# 1:      a      x     1     2
# 2:      a      y     1     1
# 3:      b      y     1     1
# 4:      b      z     3     2

Answer 3

使用dplyr（在David Arenburg＆amp; eddi的帮助下）：

library(dplyr)
dt %>% group_by(group1, group2) %>% summarise_each(funs(sum(!is.na(.))))
Source: local data table [4 x 4]
Groups: group1

  group1 group2 data1 data2
1      a      x     1     2
2      a      y     1     1
3      b      y     1     1
4      b      z     3     2

按列分组的非NA记录数

3 个答案: