使用 SE mutate 函数的动态变量名答案

【问题标题】：Dynamic variable names using SE mutate function使用 SE mutate 函数的动态变量名
【发布时间】：2018-02-27 19:42:48
【问题描述】：

我不知道如何在 for 循环中将动态变量分组。

df 是包含clm_april - clm_sept 变量的数据框，这些变量是double。我还想为新列命名。

每个月不同的组如下所示：

> groups_april
 [1] "0"       "500"     "1000"    "1500"    "2000"    "3500"   
 [7] "4500"    "5000"    "9500"    "2000000"

以下不起作用：

vector <- c("april", "may", "june", "july", "aug", "sept")

for (i in vector) {
  varname <- paste0("clm_", i)
  df <- df %>%
    mutate_(.dots = 
              cut(list(varname),
                  breaks = groups[i],
                  include.lowest = T,
                  dig.lab = 10))
}

我也试过了：

for (i in vector) {
  varname <- paste0("clm_", i)
  df <- df %>%
    mutate_(.dots = interp(~cut(list(varname),
                           breaks = groups[i],
                           include.lowest = T,
                           dig.lab = 10,  varname=as.name(varname))))
}

R 抛出以下错误：

Error in mutate_impl(.data, dots) : 
  Evaluation error: 'x' must be numeric.

你能帮我解决这个问题吗？

编辑：

# A tibble: 10 x 6
   clm_april clm_may clm_june clm_july clm_aug clm_sept
       <dbl>   <dbl>    <dbl>    <dbl>   <dbl>    <dbl>
 1         0       0        0        0     689        0
 2      2000       0     1000     1000    1000        0
 3      5000    1000     1000     1000    1500     1518
 4      1000    1069     1100     1200    2019     2000
 5       679     689     9000    10000   36681     2000
 6       800    1000     1000      657    1815     2500

【问题讨论】：

尝试给出一段数据作为示例，通过使用 dput() 或 head() 并粘贴结果。
寻求帮助时，您应该包含一个简单的reproducible example，其中包含可用于测试和验证可能解决方案的示例输入和所需输出
我编辑了这个问题。无论如何，我想我已经包含了您需要的所有必要信息。如果您需要其他东西，请告诉我。谢谢`
df %>% mutate_all(cut, breaks = group, include.lowest = T, dig.lab = 10)?
只有我一个人不明白这个问题吗？？我觉得很傻:(

标签： r dplyr dynamic-function quosure

【解决方案1】：

一种选择是使用来自purrr 包的map2_df。 map2_df 将允许用于传递实际 data.frame 的列数据。并同时传递列的名称。自定义函数find_breaks用于列名找到对应的breaks，然后调用cut。

Suppose different Groups are defined for each month is defined as:

groups_april <- c("0", "500", "1000",  "1500", "2000", "3500", "4500", "5000", "9500", "2000000")
groups_may <- c("0", "500", "1000",  "1500", "2000", "3500", "4500", "5000", "9500", "1000000")
groups_june <- c("0", "500", "1000",  "1500", "2000", "3500", "4500", "5000", "20000", "2000000")
groups_july <- c("0", "500", "1000",  "1500", "2000", "3500", "4500", "7000", "9500", "2000000")
groups_aug <- c("0", "500", "1000",  "1500", "2000", "3500", "4500", "6000", "9500", "2000000")
groups_sept <- c("0", "500", "1000",  "1500", "2000", "3500", "4500", "5000", "9500", "20000")


#Lets create a data.frame for Groups:
groups_df <- data.frame(groups_april, groups_may, groups_june, groups_july,
       groups_aug, groups_sept, stringsAsFactors = FALSE)


# Create a function to find breaks and apply cut
find_breaks <- function(x, y){
  #get the corresponding column name
  breaks_group_col <- grep(gsub("clm_", "", y), names(groups_df), value = TRUE)

  #apply cut using corresponding column from groups_df
  cut(x, breaks = groups_df[,breaks_group_col], include.lowest = TRUE, dig.lab = 10)
}

library(purrr)

# Pass df and column names of df to map2. 
map2_df(df, grep("clm_*", names(df), value = TRUE), function(.x, .y) find_breaks(.x, .y))

#Result
  clm_april   clm_may     clm_june     clm_july       clm_aug        clm_sept   
  <fctr>      <fctr>      <fctr>       <fctr>         <fctr>         <fctr>     
1 [0,500]     [0,500]     [0,500]      [0,500]        (500,1000]     [0,500]    
2 (1500,2000] [0,500]     (500,1000]   (500,1000]     (500,1000]     [0,500]    
3 (4500,5000] (500,1000]  (500,1000]   (500,1000]     (1000,1500]    (1500,2000]
4 (500,1000]  (1000,1500] (1000,1500]  (1000,1500]    (2000,3500]    (1500,2000]
5 (500,1000]  (500,1000]  (5000,20000] (9500,2000000] (9500,2000000] (1500,2000]
6 (500,1000]  (500,1000]  (500,1000]   (500,1000]     (1500,2000]    (2000,3500]

数据

df <- read.table(text = "clm_april clm_may clm_june clm_july clm_aug clm_sept
1         0       0        0        0     689        0
2      2000       0     1000     1000    1000        0
3      5000    1000     1000     1000    1500     1518
4      1000    1069     1100     1200    2019     2000
5       679     689     9000    10000   36681     2000
6       800    1000     1000      657    1815     2500", header = T, stringsAsFactors = F)

【讨论】：