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我有一个数据框，其中包含与时间段 1 相关的 3 个二进制变量和与时间段 2 相关的三个相应变量。

df <- data.frame("user" = c("a","b","c","d","e"), "item_1_time_1" = c(1,0,0,0,NA), "item_2_time_1" = c(1,1,1,0,NA), "item_3_time_1" = c(0,0,1,0,0), "item_1_time_2" = c(1,0,0,0,NA), "item_2_time_2" = c(1,0,0,0,NA), "item_3_time_2" = c(0,0,1,0,1))

df

   user item_1_time_1 item_2_time_1 item_3_time_1 item_1_time_2 item_2_time_2 item_3_time_2
1    a             1             1             0             1             1             0
2    b             0             1             0             0             0             0
3    c             0             1             1             0             0             1
4    d             0             0             1             0             0             0
5    e            NA            NA             0            NA            NA             1

我想知道观察是否在第 1 期但在第 2 期没有1给定item。此外，我想知道观察是否有任何实例，其中项目1在第 1 期而不是第 2 期。

所以理想的输出看起来像

df2 <- data.frame("user" = c("a","b","c","d","e"), "item_1_time_1" = c(1,0,0,0,NA), "item_2_time_1" = c(1,1,1,0,NA), "item_3_time_1" = c(0,0,1,1,0), "item_1_time_2" = c(1,0,0,0,NA), "item_2_time_2" = c(1,0,0,0,NA), "item_3_time_2" = c(0,0,1,0,1), "item_1_check" = c(1,1,1,1,1), "item_2_check" = c(1,0,0,1,1), "item_3_check" = c(1,1,1,0,1), item_check = c(1,0,0,0,1))

df2 

user item_1_time_1 item_2_time_1 item_3_time_1 item_1_time_2 item_2_time_2 item_3_time_2 item_1_check item_2_check item_3_check item_check
1    a             1             1             0             1             1             0            1            1            1          1
2    b             0             1             0             0             0             0            1            0            1          0
3    c             0             1             1             0             0             1            1            0            1          0
4    d             0             0             1             0             0             0            1            1            0          0
5    e            NA            NA             0            NA            NA             1            1            1            1          1

到目前为止我已经尝试过

library(tidyverse)
df2 <- df %>%
   mutate(across(ends_with('time_2'), replace_na, 0)) %>% 
   mutate(across(ends_with('time_1'), replace_na, 0)) %>% 
   mutate(item_1_check = if_else(item_1_time_1 == 1 & item_1_time_2 == 0, 0, 1),
          item_2_check = if_else(item_2_time_1 == 1 & item_2_time_2 == 0, 0, 1),
          item_3_check = if_else(item_3_time_1 == 1 & item_3_time_2 == 0, 0, 1)) %>% 
   mutate(item_check = pmin(item_1_check, item_2_check, item_3_check))

我想概括上述 mutate 调用，以便它们可以处理 n 个项目，而不仅仅是 3个。 有没有一种方法可以ends_with('check')用于最终的 mutate？变量名称没有变化，但项目编号和时间段不同。

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1 回答 1

2

一种选择是重塑为“长”格式并执行一次

library(dplyr)
library(tidyr)
df %>% 
  pivot_longer(cols = -user, names_to = c('group', '.value'), 
         names_sep="_(?=time)") %>% 
  mutate(across(starts_with('time'), replace_na, 0)) %>% 
  group_by(group) %>% 
  transmute(user, check = !(time_1 & !time_2)) %>% 
  ungroup %>% 
  group_by(user) %>%
  summarise(check = min(check), .groups = 'drop') %>% 
  right_join(df, .) %>%
  select(names(df), check)
# user item_1_time_1 item_2_time_1 item_3_time_1 item_1_time_2 item_2_time_2 item_3_time_2 check
#1    a             1             1             0             1             1             0     1
#2    b             0             1             0             0             0             0     0
#3    c             0             1             1             0             0             1     0
#4    d             0             0             0             0             0             0     1
#5    e            NA            NA             0            NA            NA             1     1

或使用base R

df$check <-  +( Reduce(`&`, lapply(split.default(replace(df[-1], 
 is.na(df[-1]), 0), sub("time_\\d+", "", names(df)[-1])), 
    function(x)  !(x[[1]] & !x[[2]]))))
于 2020-09-19T20:53:56.170 回答