Question

我需要在分析中按天重命名数据框。

names(dados) <- c("name", "day_1","Freq_1","Percent_1","day_2","Freq_2","Percent_2",
                  "day_3","Freq_3","Percent_3","day_4","Freq_4","Percent_4",
                  "day_5","Freq_5","Percent_5","day_6","Freq_6","Percent_6",
                  "day_7","Freq_7","Percent_7","day_8","Freq_8","Percent_8",
                  "day_9","Freq_9","Percent_9")

我正在分析我得到的数据是在数据框列表中，每个数据框代表一天的分析。我合并了数据框，我有列＆＃39; name＆＃39;独特的＆＃39; day_X＆＃39;，＆＃39; Freq_X＆＃39;和＆＃39; Percent_X＆＃39;将每个数据帧作为回报。

作为返回，我需要列具有以下名称：

＆＃34; name＆＃34;，＆＃34; day_1＆＃34;，＆＃34; Freq_1＆＃34;，＆＃34; Percent_1＆＃34;，＆＃34; day_2＆＃34;，＆＃ 34; Freq_2＆＃34;＆＃34; Percent_2＆＃34;＆＃34; day_3＆＃34;＆＃34; Freq_3＆＃34;＆＃34; Percent_3＆＃34;

如何分析50天？

可重复的例子：

day1 <- data.frame(name = c("jose", "mary", "julia"), freq = c(1,5,3), percent = c(40,30,20))
day2 <- data.frame(name = c("abner", "jose", "mary"), freq = c(3,5,4), percent = c(20,30,20))
day3 <- data.frame(name = c("abner", "jose", "mike"), freq = c(6,2,3), percent = c(40,30,70))
day4 <- data.frame(name = c("andre", "joseph", "ana"), freq = c(1,5,8), percent = c(40,30,20))
day5 <- data.frame(name = c("abner", "poli", "joseph"), freq = c(4,3,3), percent = c(10,30,10))

dates <- list(day1,day2,day4,day5)

data <- Reduce(function(x, y) merge(x, y, by = "name", all = TRUE), dates)

Answer 1

这是使用Get-ChildItem套件套件获得所需内容的方法。我们首先将数据放在“长”格式中 - 但添加一个日期为：

的列

tidyverse

现在，为了获得您所需的宽格式，我们需要重新格式化一些内容，如下面的代码所示。我不确定long_form <- dates %>% imap_dfr(function(x, y) dplyr::mutate(x, day_num = y))变量中应该包含哪些内容，如评论中提到的@useR，所以它不见了。如果你有一个名为day_#的变量，代码应该自动做正确的事情。

day

Answer 2

可以使用dplyr::bind_rows将列表中的所有数据帧合并到数据帧。请提供name列表，以便事先设置day1，day2等。最后，gather和spread用于转换数据。

names(dates) <- paste("day", seq_along(dates), sep = "")

library(tidyverse)

bind_rows(dates,.id = "Name") %>%
  group_by(Name) %>%
  mutate(rn = row_number()) %>% 
  ungroup() %>%
  gather(Key, value, -Name,-rn) %>%
  unite("Key", c("Key", "Name")) %>%
  spread(Key, value) %>%
  select(-rn)

<强>结果：

# # A tibble: 3 x 12
#  freq_day1 freq_day2 freq_day3 freq_day4 name_day1 name_day2 name_day3 name_day4 percent_day1 percent_day2 percent~ percent~
#   * <chr>     <chr>     <chr>     <chr>     <chr>     <chr>     <chr>     <chr>     <chr>        <chr>        <chr>    <chr>   
# 1 1         3         1         4         jose      abner     andre     abner     40           20           40       10      
# 2 5         5         5         3         mary      jose      joseph    poli      30           30           30       30      
# 3 3         4         8         3         julia     mary      ana       joseph    20           20           20       10     
#

数据：

从OP略微修改数据。我已将stringsAsFactors = FALSE参数作为data.frame的一部分，以避免mutate_at调用将factor转换为character。

day1 <- data.frame(name = c("jose", "mary", "julia"), freq = c(1,5,3), percent = c(40,30,20), stringsAsFactors = FALSE) day2 <- data.frame(name = c("abner", "jose", "mary"), freq = c(3,5,4), percent = c(20,30,20), stringsAsFactors = FALSE) day3 <- data.frame(name = c("abner", "jose", "mike"), freq = c(6,2,3), percent = c(40,30,70), stringsAsFactors = FALSE) day4 <- data.frame(name = c("andre", "joseph", "ana"), freq = c(1,5,8), percent = c(40,30,20), stringsAsFactors = FALSE) day5 <- data.frame(name = c("abner", "poli", "joseph"), freq = c(4,3,3), percent = c(10,30,10), stringsAsFactors = FALSE) dates <- list(day1,day2,day4,day5)

在R中按天重命名数据帧列

2 个答案: