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R- rbind逻辑
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Stack Overflow用户
提问于 2019-06-26 13:27:28
回答 2查看 43关注 0票数 0

我有这个数据:

代码语言:javascript
复制
source_data <- data.frame(
    "date" = c("2018-01-01", "2018-01-01", "2018-02-01", "2018-02-01"), 
    "nr" = c(0, 1, 0, 1),
    "marketing_fees" = c(500, 600, 800, 900),
    "services_paid" = c(40, 50, 10, 30),
    stringsAsFactors = F)

结果应该是这样的

代码语言:javascript
复制
result <- data.frame(
  "date" = c("2018-01-01", "2018-01-01", "2018-01-01", "2018-01-01", "2018-02-01", "2018-02-01", "2018-02-01", "2018-02-01"), 
  "nr" = c(0, 0, 1, 1, 0, 0, 1, 1),
  "income" = c(500, 40, 600, 50, 800, 10, 900, 30),
  "source" = c("marketing", "services", "marketing", "services", "marketing", "services", "marketing", "services"),
  stringsAsFactors = F)

我唯一能做的就是

代码语言:javascript
复制
result <- rbind(
  source_data %>% 
    filter(date == "2018-01-01") %>% 
    select(date, nr, income = marketing_fees) %>% 
    mutate(source = "marketing"),

  source_data %>% 
    filter(date == "2018-01-01") %>% 
    select(date, nr, income = services_paid) %>% 
    mutate(source = "services"),

  source_data %>% 
    filter(date == "2018-02-01") %>% 
    select(date, nr, income = marketing_fees) %>% 
    mutate(source = "marketing"),

  source_data %>% 
    filter(date == "2018-02-01") %>% 
    select(date, nr, income = services_paid) %>% 
    mutate(source = "services")
)

上面的代码不仅是丑陋的,有很多重复的部分,我不能再这样使用它了,我的数据文件有大约50列和很多数据。如果没有这么多重复的代码,你如何才能获得结果数据?

EN

回答 2

Stack Overflow用户

回答已采纳

发布于 2019-06-26 13:28:52

我们可以使用gather将“wide”重塑为“long”,然后separate列名只返回前缀部分

代码语言:javascript
复制
library(tidyverse)
source_data %>% 
    gather(source, income, marketing_fees:services_paid) %>% 
    separate(source, into = c('source', 'extra')) %>%
    select(-extra) %>% 
    arrange(date, nr)
#        date nr    source income
#1 2018-01-01  0 marketing    500
#2 2018-01-01  0  services     40
#3 2018-01-01  1 marketing    600
#4 2018-01-01  1  services     50
#5 2018-02-01  0 marketing    800
#6 2018-02-01  0  services     10
#7 2018-02-01  1 marketing    900
#8 2018-02-01  1  services     30
票数 1
EN

Stack Overflow用户

发布于 2019-06-26 13:47:45

代码语言:javascript
复制
library(data.table)
library(magrittr)
result2 <- melt(
  setDT(source_data), 
  id.vars = c("date", "nr"), 
  value.name = "income", 
  variable.name = "source"
)[, source := sub("_.*", "", source)][order(date, nr)]°

         date nr    source income
1: 2018-01-01  0 marketing    500
2: 2018-01-01  0  services     40
3: 2018-01-01  1 marketing    600
4: 2018-01-01  1  services     50
5: 2018-02-01  0 marketing    800
6: 2018-02-01  0  services     10
7: 2018-02-01  1 marketing    900
8: 2018-02-01  1  services     30
票数 0
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/56773987

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