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无法将base-R`rep()`与dplyr一起使用

dplyr r

2

stackinator Brenton Wiernik · 技术社区 · 7 年前

library(tidyverse)
x <- c(0, 20, 30, 58)
n <- 100
df <- data_frame(x, n) %>% 
  distinct() %>%
  filter(x >= 0 & x < n) %>%
  arrange(x) %>%
  bind_rows(data_frame(x = n)) %>%
  mutate(lag_x = lag(x)) %>%
  mutate(y = x - lag_x) %>%
  filter(!is.na(y))

rep(seq_along(df$x), df$y)

rep() 打电话?对吗?

library(tidyverse)
x <- c(0, 20, 30, 58)
n <- 100
df <- data_frame(x, n) %>% 
  distinct() %>%
  filter(x >= 0 & x < n) %>%
  arrange(x) %>%
  bind_rows(data_frame(x = n)) %>%
  mutate(lag_x = lag(x)) %>%
  mutate(y = x - lag_x) %>%
  filter(!is.na(y)) %>% 
  rep(seq_along(x), y) %>% 
  print()

函数\u list[i]出错:找不到对象“y”

1 回复 | 直到 7 年前

1

5

akrun 7 年前

因为它在外面 mutate/summarise pull 或者提取它

data_frame(x, n) %>% 
  distinct() %>%
  filter(x >= 0 & x < n) %>%
  arrange(x) %>%
  bind_rows(data_frame(x = n)) %>%
  mutate(lag_x = lag(x)) %>%
  mutate(y = x - lag_x) %>%
  filter(!is.na(y)) %>% 
  {rep(seq_along(.$x), .$y)}

或者也可以写成

data_frame(x, n) %>% 
 distinct() %>%
 filter(x >= 0 & x < n) %>%
 arrange(x) %>%
 bind_rows(data_frame(x = n)) %>%
 mutate(lag_x = lag(x)) %>%
 mutate(y = x - lag_x) %>%
 filter(!is.na(y)) %>% 
 summarise(n = list(rep(row_number(), y))) %>%
 pull(n)