Question

假设我具有以下数据框：

df <- data.frame(Order=c("1234567","1234567","1234567","456789","456789"),Stage=c("Pipeline","Proposal","Closed","Pipeline","Lost"),StageChange=c("2008-01-01","2008-01-02","2008-01-03","2008-01-10","2008-01-12"))

结果：

    head(df)
    Order    Stage StageChange
1 1234567 Pipeline  2008-01-01
2 1234567 Proposal  2008-01-02
3 1234567   Closed  2008-01-03
4  456789 Pipeline  2008-01-10
5  456789     Lost  2008-01-12

我需要拆开“阶段”（Stage）列并进入这样的数据框：

    Order   Pipeline   Proposal     Closed       Lost
1 1234567 2008-01-01 2008-01-02 2008-01-03         NA
2  456789 2008-01-10         NA         NA 2008-01-12

我阅读了文档，并尝试了dplyr和tidyr（like in this thread）的不同方法，但是我的无知正在赢得胜利。

有什么想法可以满足我的需求？

为了明确起见，我的目标是使用此数据来计算特定订单在特定阶段花费的天数。有些订单丢失，另一些则关闭（赢），这就是为什么存在“ NA”值的原因。当订单未更改到特定阶段时，也会发生同样的情况（订单可以从“流水线”转移到“丢失”，而无需对中间阶段进行任何更改）。

谢谢！

Answer 1

您可以使用tidyr::pivot_wider。这是退休功能spread

的新版本

# install.packages("tidyr")
library(tidyr)

df %>%
  pivot_wider(names_from = Stage, values_from = StageChange)

# # A tibble: 2 x 5
#   Order   Pipeline   Proposal   Closed     Lost      
#   <fct>   <fct>      <fct>      <fct>      <fct>     
# 1 1234567 2008-01-01 2008-01-02 2008-01-03 NA        
# 2 456789  2008-01-10 NA         NA         2008-01-12

Answer 2

日期将是factor类

library(tidyverse)

df_wide <- df %>%
  tidyr::pivot_wider(names_from = Stage, values_from = StageChange)

df_wide

# A tibble: 2 x 5
  Order   Pipeline   Proposal   Closed     Lost      
  <fct>   <fct>      <fct>      <fct>      <fct>     
1 1234567 2008-01-01 2008-01-02 2008-01-03 NA        
2 456789  2008-01-10 NA         NA         2008-01-12

如果您想将日期转换为Date类

df_wide_dates <- df %>%
  tidyr::pivot_wider(names_from = Stage, values_from = StageChange) %>%
  dplyr::mutate_at(., vars(Pipeline, Proposal, Closed, Lost), as.Date)

df_wide_dates

# A tibble: 2 x 5
  Order   Pipeline   Proposal   Closed     Lost      
  <fct>   <date>     <date>     <date>     <date>    
1 1234567 2008-01-01 2008-01-02 2008-01-03 NA        
2 456789  2008-01-10 NA         NA         2008-01-12

Answer 3

使用random

dplyr::spread

R：拆开包含日期的列

3 个答案: