分散列并按r中的ID进行计数

如何解决分散列并按r中的ID进行计数

我有一个因子列,名称为Lead_DataSource__c。我想将每个因子分散到一列,然后按ID在每一行中显示的因子计数来填补空白。

这是我数据框的头;

head(df)


 Id                 Lead_DataSource__c numberoflead leadduration  lasttouch firsttouch
  <chr>              <chr>                     <int> <drtn>        <chr>     <chr>     
1 0010I000026fxp6QAA NA                            1       NA days NA        NA        
2 0010I000026frM6QAI Walk in                       1   0.0000 days Walk in   Walk in   
3 0010I000026frOQQAY Walk in                       1   0.0000 days Walk in   Walk in   
4 0010I000026frsUQAQ Walk in                       3 243.9656 days Walk in   Facebook  
5 0010I000026frsUQAQ Facebook                      3 243.9656 days Walk in   Facebook  
6 0010I000026frsUQAQ Facebook                      3 243.9656 days Walk in   Facebook  

我需要这个;

            Id lastcreateddateoflead lasttouch firsttouch Facebook Walk.in <NA>
            1 0010I000026frM6QAI                 43575   Walk in    Walk in        0       1    0
            2 0010I000026frOQQAY                 43843   Walk in    Walk in        0       1    0
            3 0010I000026frsUQAQ                 43794   Walk in   Facebook        2       1    0
            4 0010I000026frsUQAQ                 43794   Walk in   Facebook        2       1    0
            5 0010I000026frsUQAQ                 43794   Walk in   Facebook        2       1    0
            6 0010I000026fsBrQAI                 43699  Facebook   Facebook        1       0    0

到目前为止,我已经使用dplyr进行了尝试,但没有得到上面想要的;

df%>%
group_by(Id,Lead_DataSource__c) %>%
 mutate(numberofleadsource=n()) %>% 
  spread(Lead_DataSource__c,numberofleadsource,fill = 0)

这是我的代码的输出;

             Id lastcreateddateoflead lasttouch firsttouch Facebook Walk.in <NA>
             1 0010I000026frM6QAI                 43575   Walk in    Walk in        0       1    0
             2 0010I000026frOQQAY                 43843   Walk in    Walk in        0       1    0
             3 0010I000026frsUQAQ                 43794   Walk in   Facebook        2       0    0
             4 0010I000026frsUQAQ                 43794   Walk in   Facebook        2       0    0
             5 0010I000026frsUQAQ                 43794   Walk in   Facebook        0       1    0
             6 0010I000026fsBrQAI                 43699  Facebook   Facebook        1       0    0

有人可以帮我解决我在这里想念的东西吗?

输入数据:

structure(list(Id = c("0010I000026fxp6QAA","0010I000026frM6QAI","0010I000026frOQQAY","0010I000026frsUQAQ","0010I000026frsUQAQ"),Lead_DataSource__c = c(NA,"Walk in","Facebook","Facebook"),numberoflead = c(1L,1L,3L,3L),leadduration = structure(c(NA,243.9656,243.9656),class = "difftime",units = "days"),lasttouch = c(NA,"Walk in"),firsttouch = c(NA,"Facebook")),row.names = c(NA,-6L),class = c("tbl_df","tbl","data.frame"))

解决方法

在这里,我使用add_count()来计算每个ID /线索源组合发生的次数,然后使用pivot_wider()进行传播。最后一行填补了数据透视表中缺失的值。

library(dplyr)
library(tidyr)
       
df %>%
  add_count(Id,Lead_DataSource__c) %>%
  mutate(tmp = 1:nrow(.)) %>%
  pivot_wider(names_from = Lead_DataSource__c,values_from = n) %>%
  select(-tmp) %>%
  group_by(Id) %>%
  mutate_at(c("NA","Walk in","Facebook"),~ifelse(any(!is.na(.)),.[!is.na(.)][1],0))
# A tibble: 6 x 8
# Groups:   Id [4]
  Id                 numberoflead leadduration  lasttouch firsttouch  `NA` `Walk in` Facebook
  <chr>                     <int> <drtn>        <chr>     <chr>      <dbl>     <dbl>    <dbl>
1 0010I000026fxp6QAA            1       NA days NA        NA             1         0        0
2 0010I000026frM6QAI            1   0.0000 days Walk in   Walk in        0         1        0
3 0010I000026frOQQAY            1   0.0000 days Walk in   Walk in        0         1        0
4 0010I000026frsUQAQ            3 243.9656 days Walk in   Facebook       0         1        2
5 0010I000026frsUQAQ            3 243.9656 days Walk in   Facebook       0         1        2
6 0010I000026frsUQAQ            3 243.9656 days Walk in   Facebook       0         1        2

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