如何找到满足条件的群的第一个元素

first_equal_to = function(x, value)
    (x == value) & (cumsum(x == value) == 1)

所以

tbl %>% group_by(group) %>% mutate(first = first_equal_to(var, 1))

(似乎应该将其作为逻辑向量,因为这是列所表示的内容)。

first_equal_to2 = function(x, value) {
    result = logical(length(x))
    result[match(value, x)] = TRUE
    result
}

library(dplyr)

df$first = NULL

df %>%
  group_by(group) %>%
  mutate(first = as.numeric(row_number() == min(row_number()[var == 1]))) %>%
  ungroup()

# # A tibble: 26 x 3
#   group   var first
#   <int> <int> <dbl>
# 1    17    74     0
# 2    17    49     0
# 3    17     1     1
# 4    18    74     0
# 5    18     1     1
# 6    18    49     0
# 7    18    61     0
# 8    19    49     0
# 9    19     1     1
# 10   19     5     0
# # ... with 16 more rows

其思想是在 var =1,每组内。

这将返回一些警告,因为在某些组中没有 变量 =1例。

另一种选择是:

library(dplyr)

df$first = NULL

# create row id
df$id = seq_along(df$group)

df %>%
  filter(var == 1) %>%                         # keep cases where var = 1
  distinct(group, .keep_all = T) %>%           # keep distinct cases based on group
  mutate(first = 1) %>%                        # create first column
  right_join(df, by=c("id","group","var")) %>% # join back original dataset
  mutate(first = coalesce(first, 0)) %>%       # replace NAs with 0
  select(-id)                                  # remove row id

# # A tibble: 26 x 3
#   group   var first
#   <int> <int> <dbl>
# 1    17    74     0
# 2    17    49     0
# 3    17     1     1
# 4    18    74     0
# 5    18     1     1
# 6    18    49     0
# 7    18    61     0
# 8    19    49     0
# 9    19     1     1
#10    19     5     0
# # ... with 16 more rows

我们可以使用所示的表达式 first :

DF %>% 
  group_by(group) %>% 
  mutate(first = { var == 1 } %>% { . * !duplicated(.) } ) %>%
  ungroup

给:

# A tibble: 26 x 3
   group   var first
   <int> <int> <int>
 1    17    74     0
 2    17    49     0
 3    17     1     1
 4    18    74     0
 5    18     1     1
 6    18    49     0
 7    18    61     0
 8    19    49     0
 9    19     1     1
10    19     5     0
# ... with 16 more rows