如何有效地进行数据争吵以自动化 Excel 流程？答案

【问题标题】：How to data wrangle efficiently to automatize the excel process?如何有效地进行数据争吵以自动化 Excel 流程？
【发布时间】：2019-09-20 10:42:27
【问题描述】：

我们通过 excel 执行的宏太多了，但是 我们可以使用 R 实现自动化吗？。

我还不确定我还不是 R 专家需要什么。像聚集、传播或重塑这样的功能没有多大意义…… 可能 stringr 和 regrex 可能会起作用，但不够熟练，无法尝试。

基本上，

我们必须从“列表”列中删除 ID，例如 beet_root、apple（通常以字母开头，除了少数已知的例外，例如 ot_p）。当然，我们也可以完全删除该行。
创建新列“名称”，然后附加对应 ID 例如列表值的橙色 ID：9734,75R4 而不是与葡萄 ID 对应的 123/R90

请找到reprex（简化的excel流程）

library(tidyverse)



> input
        list other_col1 other_col2
1  beet_root          a          u
2    123/92R          b          b
3    123/92R          c          q
4    10.1233          d          p
5      9.485          e          e
6       ot_p          f          f
7      apple          g          b
8    12X0893          z          z
9      123fg          h          8
10     038q4          i          i
11    orange          j          j
12      9734          k          9
13      75R4          l          l
14    grapes          m          m
15   123/R90          n          5
16   90X83.6          o          o


> expected_output
      list other_col1 other_col2     names
1  123/92R          b          b beet_root
2  123/92R          c          q beet_root
3  10.1233          d          p beet_root
4    9.485          e          e beet_root
5     ot_p          f          f beet_root
6  12X0893          z          z     apple
7    123fg          h          8     apple
8    038q4          i          i     apple
9     9734          k          9    orange
10    75R4          l          l    orange
11 123/R90          n          5    grapes
12 90X83.6          o          o    grapes

数据：

# Actual data 
input = data.frame("list" = c("beet_root","123/92R","123/92R","10.1233","9.485","ot_p",
                            "apple","12X0893","123fg","038q4",
                            "orange","9734","75R4",
                           "grapes", "123/R90","90X83.6"),
                "other_col1" = c("a","b","c","d","e","f","g","z",
                                "h","i","j","k","l","m","n","o"),
                "other_col2" = c("u","b","q","p","e","f","b","z",
                                "8","i","j","9","l","m","5","o"))
expected_output = data.frame("list" = c("123/92R","123/92R","10.1233","9.485","ot_p",
                                      "12X0893","123fg","038q4",
                                      "9734","75R4",
                                      "123/R90","90X83.6"),
                           "other_col1" = c("b","c","d","e",
                                            "f","z","h","i",
                                            "k","l",
                                            "n","o"
                                            ),
                           "other_col2" = c("b","q","p","e",
                                            "f","z","8","i",
                                            "9","l",
                                            "5","o"),
                           "names" = c("beet_root","beet_root","beet_root","beet_root","beet_root",
                                       "apple","apple","apple",
                                       "orange","orange",
                                       "grapes","grapes"))

【问题讨论】：

听起来像是合并？
真的吗？只有一张表作为输入和输出
问题是输入和预期输出是隐藏的，这使得它有点难以理解。

标签： r string dplyr

【解决方案1】：

如果它遵循特定模式（以数字开头或为"ot_p"）或具有NA，我们创建一个新列names，其值来自list。我们将fill 的NA 值向下names，然后对filter 行使用相同的正则表达式。

library(dplyr)

input %>%
  mutate(names = ifelse(!grepl("^[0-9]|ot_p", list), list, NA)) %>% 
  tidyr::fill(names) %>%
  filter(grepl("^[0-9]|ot_p", list))

#      list other_col1 other_col2     names
#1  123/92R          b          b beet_root
#2  123/92R          c          q beet_root
#3  10.1233          d          p beet_root
#4    9.485          e          e beet_root
#5     ot_p          f          f beet_root
#6  12X0893          z          z     apple
#7    123fg          h          8     apple
#8    038q4          i          i     apple
#9     9734          k          9    orange
#10    75R4          l          l    orange
#11 123/R90          n          5    grapes
#12 90X83.6          o          o    grapes

首先运行input[] <- lapply(input, as.character) 将factors 转换为字符。

【讨论】：

【解决方案2】：

tidyverse 的选项

library(tidyverse0
input %>% 
  mutate(names =  case_when(str_detect(list, "^([0-9]|ot_p)") ~ 
            NA_character_, TRUE ~ as.character(list) )) %>%
  fill(names) %>%
  filter(str_detect(list, "^[0-9]|ot_p"))

【讨论】：