Question

我有一个datframe，我想每天甚至在三天内计算变化百分比，但是当我这样做时，结果似乎并不正确。

ads <- data.frame(ad = c(ad1, ad1, ad1, ad1, ad2, ad2, ad2, ad3, ad3, ad3), 
                  date = c("11-10", "11-11", "11-12", "11-13", "11-10", "11-11", "11-12", "11-10", "11-11", "11-12"), 
                  likes = c(20, 30, 18, 5, 34, 68, 55, 44, 33, 20),
                  comments = c(21, 22, 10, 1, 10, 43, 24, 34, 21, 11))

所以我有这个：

daily_pct <- function(x) x/lag(x)
three_pct <- function(x) x/lag(x ,k = 3)

daily_pct_change <- ads %>%
     mutate_each(funs(daily_pct), c(likes,comments))

three_pct_change <- ads %>% 
     mutate_each(funs(three_pct), c(likes, comments))

我这样做正确吗？我也不知道如何让三天一班的工作。谢谢！

Answer 1

您可以尝试：

df %>% 
  mutate_at(.vars = vars(dplyr::matches("(likes)|(comments)")), 
            funs(daily_change = ./lag(.)*100,
                 three_day_change = ./lag(., 3)*100))

类似地，如果您不需要ad和date变量：

df %>% 
  select(likes, comments) %>% 
  mutate_all(funs(daily_change = ./lag(.)*100,
                 three_day_change = ./lag(., 3)*100))

或者如果您需要它们：

df %>% 
  select(likes, comments) %>% 
  mutate_all(funs(daily_change = ./lag(.)*100,
                 three_day_change = ./lag(., 3)*100)) %>% 
  rowid_to_column() %>% 
  left_join(df %>% rowid_to_column() %>% select(rowid, ad, date), by = c("rowid" = "rowid")) %>%
  select(-rowid)

此外，您可以通过对原始代码进行少量修改来获得相同的结果：

daily_pct <- function(x) x/lag(x)*100
three_pct <- function(x) x/lag(x, 3)*100

df %>% 
  mutate_at(.vars = vars(dplyr::matches("(likes)|(comments)")), 
            funs(daily_change = daily_pct,
                 three_day_change = three_pct))

如何计算每日变化百分比和三天变化百分比R？

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