R 基于多种条件重塑数据帧

我想确定在时间t期间在同一地点和同一id人员执行的活动。变量

wher

表示时间步长，并记录活动在时间t发生的位置。with参数记录在时间t执行活动的人员。我想知道在时间t期间，在同一地点和同一个人进行的基于性别的常见活动。不寻常的活动和在不同地点与不同人员执行的活动我替换为0

输入

输出：

id  t1  t2  t3  t4  
12   0   0  12  0  

18个时间步的样本数据

：

structure(list(serial = c(11011202, 11011202), DMSex = c(1, 2
), act1_1 = c(110, 110), act1_2 = c(110, 110), act1_3 = c(110, 
110), act1_4 = c(110, 110), act1_5 = c(110, 110), act1_6 = c(110, 
110), act1_7 = c(110, 110), act1_8 = c(110, 110), act1_9 = c(110, 
110), act1_10 = c(110, 110), act1_11 = c(110, 110), act1_12 = c(8219, 
110), act1_13 = c(310, 110), act1_14 = c(3210, 110), act1_15 = c(3110, 
110), act1_16 = c(7241, 110), act1_17 = c(210, 110), act1_18 = c(3819, 
110), wher_1 = c(11, 11), wher_2 = c(11, 11), wher_3 = c(11, 
11), wher_4 = c(11, 11), wher_5 = c(11, 11), wher_6 = c(11, 11
), wher_7 = c(11, 11), wher_8 = c(11, 11), wher_9 = c(11, 11), 
    wher_10 = c(11, 11), wher_11 = c(11, 11), wher_12 = c(11, 
    11), wher_13 = c(11, 11), wher_14 = c(11, 11), wher_15 = c(11, 
    11), wher_16 = c(11, 11), wher_17 = c(11, 11), wher_18 = c(11, 
    11), wit4_1 = c(0, 0), wit4_2 = c(0, 0), wit4_3 = c(0, 0), 
    wit4_4 = c(0, 0), wit4_5 = c(0, 0), wit4_6 = c(0, 0), wit4_7 = c(0, 
    0), wit4_8 = c(0, 0), wit4_9 = c(0, 0), wit4_10 = c(0, 0), 
    wit4_11 = c(0, 0), wit4_12 = c(0, 0), wit4_13 = c(0, 0), 
    wit4_14 = c(0, 0), wit4_15 = c(0, 0), wit4_16 = c(0, 0), 
    wit4_17 = c(0, 0), wit4_18 = c(0, 0)), row.names = 1:2, class = "data.frame")

---

其中，

act1

是

t

wit4

是

wit

和

wher

是

wher
一种结合
dplyr
和
purrr
的解决方案可以是：

map(.x = as.character(1:4),
    ~ df %>%
     select(id, ends_with(.x)) %>%
     group_by(id) %>%
     mutate_at(vars(matches("^wher|^wit")), ~ all(. == first(.))) %>%
     ungroup() %>%
     mutate(cond = rowSums(select(.,  matches("^wher|^wit"))) == 2) %>%
     group_by(id) %>%
     mutate_at(vars(starts_with("t")), ~ all(. == first(.)) * cond * .) %>%
     ungroup() %>%
     select(starts_with("t"))) %>%
 bind_cols(df %>%
            select(id)) %>%
 group_by(id) %>%
 summarise_all(first)

     id    t1    t2    t3    t4
  <int> <int> <int> <int> <int>
1    12     0     0    12     0

谢谢，如果你有时间，请你解释一下你的编码方式。提供了一些解释：）你的意思是你的输出不是从t1到t100排序的？非常感谢你的时间，所以我的意思是，如果我使用求幂运算符，我不会得到100个时间步，我得到了更多，但我不知道如何减少这个。所以你的问题是你不知道最终会有多少对？您是否在寻找一种动态解决方案，它可以从1到实际的对数（比如说99）进行转换？
map(.x = as.character(1:4),
    ~ df %>%
     select(id, ends_with(.x)) %>%
     group_by(id) %>%
     mutate_at(vars(matches("^wher|^wit")), ~ all(. == first(.))) %>%
     ungroup() %>%
     mutate(cond = rowSums(select(.,  matches("^wher|^wit"))) == 2) %>%
     group_by(id) %>%
     mutate_at(vars(starts_with("t")), ~ all(. == first(.)) * cond * .) %>%
     ungroup() %>%
     select(starts_with("t"))) %>%
 bind_cols(df %>%
            select(id)) %>%
 group_by(id) %>%
 summarise_all(first)

     id    t1    t2    t3    t4
  <int> <int> <int> <int> <int>
1    12     0     0    12     0

map(.x = str_extract(names(df)[grepl("^act", names(df))], "_.*+$"),
    ~ df %>%
     select(serial, ends_with(.x)) %>%
     group_by(serial) %>%
     mutate_at(vars(matches("^wher|^wit")), ~ all(. == first(.))) %>%
     ungroup() %>%
     mutate(cond = rowSums(select(.,  matches("^wher|^wit"))) == 2) %>%
     group_by(serial) %>%
     mutate_at(vars(starts_with("act")), ~ all(. == first(.)) * cond * .) %>%
     ungroup() %>%
     select(starts_with("act"))) %>%
 bind_cols(df %>%
            select(serial)) %>%
 group_by(serial) %>%
 summarise_all(first)

  serial act1_1 act1_2 act1_3 act1_4 act1_5 act1_6 act1_7 act1_8 act1_9 act1_10 act1_11 act1_12
   <dbl>  <dbl>  <dbl>  <dbl>  <dbl>  <dbl>  <dbl>  <dbl>  <dbl>  <dbl>   <dbl>   <dbl>   <dbl>
1 1.10e7    110    110    110    110    110    110    110    110    110     110     110       0
# … with 6 more variables: act1_13 <dbl>, act1_14 <dbl>, act1_15 <dbl>, act1_16 <dbl>,
#   act1_17 <dbl>, act1_18 <dbl>