dplyr 变异以创建新列并修改数据中的所有列
[英]dplyr mutate across to create a new column and modify all columns in the data
问题描述
I have some data which looks like:
我有一些数据看起来像：
column1 column2 column3 column4 column5 column6 column7 column8 column9 column10
<chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <lgl>
1 company42 NA NA NA NA NA NA NA NA NA
2 company105 NA €315k NA NA NA NA Mar 2015 NA NA
3 company23 NA NA NA NA NA NA NA NA NA
4 company70 NA €570 NA NA NA NA Apr 2016 NA NA
I want to do two things to it.我想做两件事。
- Extract all of the EUR signs and put them into a column "currency" - each row contains currency data where the currency is unique across the columns but it can change down the rows.提取所有 EUR 符号并将它们放入“货币”列中 - 每行包含货币数据，其中货币在各列中是唯一的，但它可以向下更改行。
- Convert all of the "K" to "000" and all of the "M" to "000000".将所有“K”转换为“000”，将所有“M”转换为“000000”。
Expected output:预期输出：
column1 column2 column3 column4 column5 column6 column7 column8 column9 column10 column 11
<chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <lgl> <chr>
1 company42 NA NA NA NA NA NA NA NA NA NA
2 company105 NA 315000 NA NA NA NA Mar 2015 NA NA €
3 company23 NA NA NA NA NA NA NA NA NA NA
4 company70 NA 570 NA NA NA NA Apr 2016 NA NA €
Where the new column 11 has been added and the € removed from column 3 and finally the "K" has been converted to "000".添加了新的第 11 列并从第 3 列中删除了 €，最后将“K”转换为“000”。
Data:数据：
data <- structure(list(column1 = c("company42", "company105", "company23",
"company70", "company77", "company51", "company20", "company17",
"company78", "company80", "company39", "company37", "company101",
"company61", "company104", "company41", "company88", "company131",
"company102", "company45"), column2 = c(NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, "\20060k", NA, NA, NA, NA, NA, NA, NA
), column3 = c(NA, "\200315k", NA, "\200570", NA, NA, NA, NA,
NA, "$1.05M", NA, NA, "\200177k", NA, NA, NA, "\20070k", NA,
NA, "\200223k"), column4 = c(NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_), column5 = c(NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_), column6 = c(NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, "\200653k", NA,
NA, NA, NA, NA, NA), column7 = c(NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_), column8 = c(NA,
"Mar 2015", NA, "Apr 2016", NA, NA, NA, NA, NA, "Sep 2012", NA,
NA, NA, NA, NA, NA, "Jul 2014", NA, NA, "May 2016"), column9 = c(NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_),
column10 = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA)), row.names = c(NA, -20L
), class = c("tbl_df", "tbl", "data.frame"))
2 个解决方案
解决方案1
2 2020-10-27 13:05:35
Try this approach, maybe is not the most optimal but it can address you to a new path when modifying it:试试这个方法，也许不是最理想的，但它可以在修改它时为您找到一条新路径：
library(tidyverse)
#Code 1
data2 <- data %>% mutate(across(column2:column7,~ifelse(grepl('k',.),gsub('k','000',.),
ifelse(grepl('M',.),gsub('M','000000',.),.))))
#Code 2
data2$Currency <- apply(data2[,2:7],1,
function(x) trimws(gsub('NA','',
paste0(gsub("[[:digit:]]", "", x),
collapse = ',')),whitespace = ','))
data2$Currency <- gsub('\\.','',data2$Currency)
#Code 3
data3 <- data2 %>% mutate(across(column2:column7,~gsub("[[:punct:]]", "", .)))
data3 <- data3 %>% mutate(across(column2:column7,~gsub("€", "", .)))
data3 <- data3 %>% mutate(across(column2:column7,as.numeric))
Output:输出：
# A tibble: 20 x 11
column1 column2 column3 column4 column5 column6 column7 column8 column9 column10 Currency
<chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <chr> <chr> <lgl> <chr>
1 company42 NA NA NA NA NA NA NA NA NA ""
2 company105 NA 315000 NA NA NA NA Mar 2015 NA NA "€"
3 company23 NA NA NA NA NA NA NA NA NA ""
4 company70 NA 570 NA NA NA NA Apr 2016 NA NA "€"
5 company77 NA NA NA NA NA NA NA NA NA ""
6 company51 NA NA NA NA NA NA NA NA NA ""
7 company20 NA NA NA NA NA NA NA NA NA ""
8 company17 NA NA NA NA NA NA NA NA NA ""
9 company78 NA NA NA NA NA NA NA NA NA ""
10 company80 NA 105000000 NA NA NA NA Sep 2012 NA NA "$"
11 company39 NA NA NA NA NA NA NA NA NA ""
12 company37 NA NA NA NA NA NA NA NA NA ""
13 company101 60000 177000 NA NA NA NA NA NA NA "€,€"
14 company61 NA NA NA NA 653000 NA NA NA NA "€"
15 company104 NA NA NA NA NA NA NA NA NA ""
16 company41 NA NA NA NA NA NA NA NA NA ""
17 company88 NA 70000 NA NA NA NA Jul 2014 NA NA "€"
18 company131 NA NA NA NA NA NA NA NA NA ""
19 company102 NA NA NA NA NA NA NA NA NA ""
20 company45 NA 223000 NA NA NA NA May 2016 NA NA "€"
解决方案2
1 2020-10-27 13:23:55
Does this work:这是否有效：
> data %>% rowwise() %>% mutate(column11 = case_when(any(grepl('\u20AC', c_across(column2:column10))) ~ '\u20AC',
+ any(grepl('\\$', c_across(column2:column10))) ~ '$', TRUE ~ NA_character_)) %>%
+ mutate(across(2:10, ~ str_remove_all(., '\\$|\u20AC'))) %>% mutate(across(2:10, ~ case_when(grepl('k$',.) ~ parse_number(.)*1000,
+ grepl('M$',.) ~ parse_number(.)*1000000,
+ TRUE ~ NA_real_)))
# A tibble: 20 x 11
# Rowwise:
column1 column2 column3 column4 column5 column6 column7 column8 column9 column10 column11
<chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <chr>
1 company42 NA NA NA NA NA NA NA NA NA NA
2 company105 NA 315000 NA NA NA NA NA NA NA €
3 company23 NA NA NA NA NA NA NA NA NA NA
4 company70 NA NA NA NA NA NA NA NA NA €
5 company77 NA NA NA NA NA NA NA NA NA NA
6 company51 NA NA NA NA NA NA NA NA NA NA
7 company20 NA NA NA NA NA NA NA NA NA NA
8 company17 NA NA NA NA NA NA NA NA NA NA
9 company78 NA NA NA NA NA NA NA NA NA NA
10 company80 NA 1050000 NA NA NA NA NA NA NA $
11 company39 NA NA NA NA NA NA NA NA NA NA
12 company37 NA NA NA NA NA NA NA NA NA NA
13 company101 60000 177000 NA NA NA NA NA NA NA €
14 company61 NA NA NA NA 653000 NA NA NA NA €
15 company104 NA NA NA NA NA NA NA NA NA NA
16 company41 NA NA NA NA NA NA NA NA NA NA
17 company88 NA 70000 NA NA NA NA NA NA NA €
18 company131 NA NA NA NA NA NA NA NA NA NA
19 company102 NA NA NA NA NA NA NA NA NA NA
20 company45 NA 223000 NA NA NA NA NA NA NA €
>
声明:本站的技术帖子网页，遵循CC BY-SA 4.0协议，如果您需要转载，请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.
相关问题
使用 dplyr 在除某些列之外的所有列中进行变异 - Mutate across all but some columns using dplyr
使用dplyr mutate根据列名向量创建新列 - use dplyr mutate to create new columns based on a vector of column names
为什么在 dplyr 中 mutate + across 创建末尾带有“[,1]”的列？ - Why mutate + across in dplyr create columns with "[,1]" at the end?
如何使用 dplyr 变异从将一列输入到返回列表的 function 中创建新列？ - How to use dplyr mutate to create new columns from inputting one column into a function that returns a list?
R dplyr使用自定义函数变异多列来创建新列 - R dplyr mutate multiple columns using custom function to create new column
dplyr中的mutate_each：使用其他具有相同文本的列的平均行值创建新列 - mutate_each in dplyr: create new column with the mean row values of other columns with some text in common
如何使用 purrr 中的 map 和 dplyr::mutate 根据列对创建多个新列 - How to use map from purrr with dplyr::mutate to create multiple new columns based on column pairs
R dataframe 使用cross / all_of / mutate_if从现有列创建多个新列 - R dataframe create mulitple new columns from existing columns using across / all_of / mutate_if
使用dplyr :: mutate将函数的参数化应用于单个数据框列，然后将结果保存到新列？ - Using dplyr::mutate to apply parameterizations of a function to a single data frame column, then save the results to new columns?
根据 R 中数据框中所有其他列中的字符串值，使用 dplyr 创建一个新列 - Create a new column using dplyr based on string values in all other columns in a data frame in R
相关标签