Question

案例1：输入

level = tonumber(level)

预期输出

ST_DATE ND_DATE LO_NO   ACTV_CODE   ACTV_AMT    AB_NO   FEATURE_CODE    L_NU    
7/27/16 7/27/16 265       O          15          1      INTEREST        855          
7/27/16 7/27/16 265       O          14          1      INTEREST 855

案例2：输入（我的代码适用于case2但是为case1抛出错误）

ST_DATE ND_DATE LO_NO   ACTV_CODE   ACTV_AMT    AB_NO   FEATURE_INTEREST     L_NU   
7/27/16 7/27/16 265      O           29          1             2             855

获得以下错误

ST_DATE ND_DATE LO_NO   ACTV_CODE   ACTV_AMT    AB_NO   FEATURE_CODE    L_NU    
7/27/16 7/27/16 265   O          15       1     INTEREST        855          
7/27/16 7/27/16 265   O          14       1     INSTALLMENT   855    

ST_DATE ND_DATE LO_NO   ACTV_CODE   ACTV_AMT    AB_NO   INTEREST INSTALLMENT     L_NU   
7/27/16 7/27/16 265      O           29           1      1          1           855


install_cntdup_less1 <- install_BAN %>% 
   group_by(AB_NO,LO_NO,L_NU)%>% 
   mutate(ACTV_AMT = sum(ACTV_AMT),ftr=sum(unique(!is.na(FEATURE_CODE))))%>%  
   spread(FEATURE_CODE,ftr,fill = 0)%>%
   slice(which.min(as.Date(ST_DATE, '%Y/%m/%d')))%>% 
   slice(which.max(as.Date(ND_DATE, '%Y/%m/%d')))

尝试引入如下所述的唯一ID链接，但它正在弄乱我的输出 Spread with duplicate identifiers (using tidyverse and %>%)

Error: Duplicate identifiers for rows (29424, 29425, 29426), (7415, 7416), (30120, 30121)

我无法理解做什么可以帮助我解决错误。这似乎是重复的问题，但它不是

Answer 1

这会有所帮助：

library(dplyr)
library(tidyr)

# example data
dt = read.table(text = "
                ST_DATE ND_DATE LO_NO   ACTV_CODE   ACTV_AMT    AB_NO   FEATURE_CODE    L_NU    
                7/27/16 7/27/16 265       O          15          1      INTEREST        855          
                7/27/16 7/27/16 265       OO          14          1      INTEREST        855
                7/27/16 7/27/16 265       O          15          1      OTHER        855          
                7/27/16 7/27/16 265       OO          14          1      OTHER        855 
                ", header=T, stringsAsFactors = F)

dt %>%
  group_by(AB_NO,LO_NO,L_NU)%>% 
  mutate(ACTV_AMT = sum(ACTV_AMT),
         ST_DATE = min(ST_DATE),
         ND_DATE = max(ND_DATE)) %>%
  ungroup() %>%
  mutate(id = row_number(),
         FEATURE_CODE = paste0("FEATURE_", FEATURE_CODE),
         ACTV_CODE = paste0("ACTV_", ACTV_CODE),
         count_FEATURE = 1,
         count_ACTV = 1) %>%
  spread(FEATURE_CODE, count_FEATURE) %>%
  spread(ACTV_CODE, count_ACTV) %>%
  select(-id) %>%
  group_by(ST_DATE, ND_DATE, LO_NO, ACTV_AMT, AB_NO, L_NU) %>%
  summarise_all(sum, na.rm=T) %>%
  ungroup()

# # A tibble: 1 x 10
#     ST_DATE ND_DATE LO_NO ACTV_AMT AB_NO  L_NU FEATURE_INTEREST FEATURE_OTHER ACTV_O ACTV_OO
#       <chr>   <chr> <int>    <int> <int> <int>            <dbl>         <dbl>  <dbl>   <dbl>
#   1 7/27/16 7/27/16   265       58     1   855                2             2      2       2

采用唯一计数并对R中的每个唯一值求和

1 个答案: