R中的左移列 [英] Left shift columns in R

问题描述

我有一个像这样的数据集

I have a data set like this

temp <- structure(list(col_1 = c("", "P9603", "", "", "11040", 
"80053"), col_2 = c("84484", "80061", "", "80061", "A0428", "85025"
), col_3 = c("V2632", "82310", "", "", "", "86357"), col_4 = c("J1170", 
"84305", "62311", "80061", "", ""), col_5 = c("", "86708", "J0690", 
"", "", "")), .Names = c("col_1", "col_2", "col_3", "col_4", 
"col_5"), class = c("data.table", "data.frame"))

   col_1 col_2 col_3 col_4 col_5
1:       84484 V2632 J1170      
2: P9603 80061 82310 84305 86708
3:                   62311 J0690
4:       80061       80061                        
5: 11040 A0428                  
6: 80053 85025 86357 

是否可以像这样移动列

   col_1 col_2 col_3 col_4 col_5
1: 84484 V2632 J1170              #LEFT SHIFT 1
2: P9603 80061 82310 84305 86708  #NO CHANGE
3: 62311 J0690                    #LEFT SHIFT 3
4: 80061 80061                    #LEFT SHIFT 1 FOR FIRST ITEM, 
                                  #LEFT SHIFT 2 FOR 2ND ITEM 
5: 11040 A0428                    #NO CHANGE
6: 80053 85025 86357              #NO CHANGE

我将列左移，如果值左侧为空

I am shifting columns left, if the value on left is empty

推荐答案

以下是使用 data.table 的选项。按行顺序分组，不列出 data.table的子集（ .SD ）， un ==''）> order ，转换为 list 然后在删除 grp列之后，使用原始列名称设置名称。

Here is an option using data.table. Grouped by the sequence of rows, unlist the Subset of data.table (.SD), order by the logical vector (un==''), convert to list and then set the names with the original column names after removing the 'grp' column

setnames(temp[, {un <- unlist(.SD); as.list(un[order(un=='')])},
    .(grp = 1:nrow(temp))][, grp := NULL], names(temp))[]
#  col_1 col_2 col_3 col_4 col_5
#1: 84484 V2632 J1170            
#2: P9603 80061 82310 84305 86708
#3: 62311 J0690                  
#4: 80061 80061                  
#5: 11040 A0428                  
#6: 80053 85025 86357       

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或者另一个选择是在创建序列列后将融化到长格式，然后 dcast 转换为宽格式

---

Or another option is to melt into long format after creating a sequence column, then dcast it to wide format

dcast(melt(temp[, n := seq_len(.N)], id.var = 'n')[order(n, value == ''),
     .(value, variable = names(temp)[1:5]), n], n ~ variable)[, n := NULL][]

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