Question

如何将以下格式转换为循环？

alphateam_df <- dplyr::filter(data, grepl("alphateam", Winners))
alphateam_mean <- mean(alphateam_df$growth)
betateam_df <- dplyr::filter(data, grepl("betateam", Winners))
betateam_mean <- mean(betateam_df$growth)

在其他语言中，我会写一些类似的东西：

for (team in c("alphateam","betateam")) {
   {team}_df <- dplyr::filter(data, grepl({team}, Winners))
   {team}_mean <- mean({team}_df$growth)
}

通过一些搜索我得到了以下内容，但卡在了平均（）位上。还想知道它是否需要这么复杂？

for (team in c("alphateam", "betateam")) {
  assign(paste(team, "_df", sep = ''), dplyr::filter(full, grepl(team, Winners)))
  assign(paste(team, "_mean", sep = ''), mean(...)
}

Answer 1

正如@ mt1022已经提到的那样，您可能需要考虑不使用for循环。以下示例使用group_by中的summarise和dplyr函数来获取每组获胜者的平均值。

suppressPackageStartupMessages(library(dplyr))

# example data
df <- tibble(Winners = c("alphateam", "betateam", "alphateam", "alphateam",
                         "betateam", "betateam"), 
             growth = c(4, 7, 2, 12, 7, 10))

# mean for each Winners group
df %>%
  group_by(Winners) %>%
  summarise(mean_growth = mean(growth))
#> # A tibble: 2 x 2
#>     Winners mean_growth
#>       <chr>       <dbl>
#> 1 alphateam           6
#> 2  betateam           8

有关详细信息，请参阅summarise示例http://dplyr.tidyverse.org/reference/summarise.html

在循环中引用动态变量

1 个答案: