添加新列并根据另一列中的值在列中插入值
[英]Add new columns and insert values in columns based on value in another column
问题描述
I have a R dataframe data1 as below: 
我有一个R dataframe data1 ，如下所示：
prodID storeID Term Exit
1 1001 5 0
1 1002 4 1
1 1003 3 1
1 1004 5 0
2 1001 4 1
2 1002 3 1
2 1003 5 0
3 1001 4 1
3 1002 3 1
3 1003 5 0
4 1001 4 1
4 1002 3 1
5 1001 5 0
5 1002 4 1
5 1003 3 1
This is of course highly simplified format of my real data which runs into around 3 million rows. 当然，这是我的真实数据的高度简化格式，大约有300万行。 I have to do the following: 我必须执行以下操作：
- Based on max value in
Termcolumn, insert that many columns indata1withNAvalues.根据Term列中的最大值，在data1插入具有NA值的许多列。 Column names should beWeek1,Week2,Week3, etc列名应Week1，Week2，Week3等等 - For each row, fill the new columns with 0, 1 or
NAusing these rules: 1)IfTermis 5, then insert 0 inWeek1,Week2, uptoWeek4and 1 inWeek52) IfTermis 4 then insert 0 inWeek1,Week2, andWeek3, 1 inWeek4and keepNAinWeek5.对于每一行，填充新的列与0,1或NA使用这些规则：1）如果Term是5，则在插入0Week1，Week2，高达Week4在和1Week52）如果Term是4，则在插入0Week1，Week2和Week3，1Week4并保持NA在Week5。 And so on....等等....
The final output should look like: 最终输出应如下所示：
prodID storeID Term Exit Week1 Week2 Week3 Week4 Week5
1 1001 5 0 0 0 0 0 1
1 1002 4 1 0 0 0 1 NA
1 1003 3 1 0 0 1 NA NA
1 1004 5 0 0 0 0 0 1
2 1001 4 1 0 0 0 1 NA
2 1002 3 1 0 0 1 NA NA
2 1003 5 0 0 0 0 0 1
3 1001 4 1 0 0 0 1 NA
3 1002 3 1 0 0 1 NA NA
3 1003 5 0 0 0 0 0 1
4 1001 4 1 0 0 0 1 NA
4 1002 3 1 0 0 1 NA NA
5 1001 5 0 0 0 0 0 1
5 1002 4 1 0 0 0 1 NA
5 1003 3 1 0 0 1 NA NA
This is what I tried: 这是我尝试的：
variant <- c("Week1","Week2","Week3","Week4","Week5")
data1[variant] <- NA
for (i in 1:length(data1$prodID)){
data1$Week1 <- ifelse(data1$Term==1,1,0)
data1$Week2 <- ifelse(data1$Term==2,1,0)
data1$Week3 <- ifelse(data1$Term==3,1,0)
data1$Week4 <- ifelse(data1$Term==4,1,0)
data1$Week5 <- ifelse(data1$Term==5,1,0)
}
This doesn't help me populate NA in the required cells. 这无助于我在所需的单元格中填充NA 。 I would like to retain the NA values because I am going to do a wide to long data transformation on the data frame later on. 我想保留NA值，因为稍后我将在数据帧上进行从宽到长的数据转换。 And I know the above approach is not feasible in my huge dataset. 而且我知道上述方法在我的庞大数据集中不可行。 Any suggestions are most welcome. 任何建议都是最欢迎的。
3 个解决方案
解决方案1
3 已采纳 2018-06-25 18:09:07
Here is one idea. 这是一个主意。 We can create the content you need and then split the columns. 我们可以创建所需的内容，然后拆分列。
library(dplyr)
library(data.table)
library(splitstackshape)
dat2 <- dat %>%
mutate(Week = case_when(
Term == 5 ~"0,0,0,0,1",
Term == 4 ~"0,0,0,1,NA",
Term == 3 ~"0,0,1,NA,NA",
Term == 2 ~"0,1,NA,NA,NA",
Term == 1 ~"1,NA,NA,NA,NA"
)) %>%
cSplit(splitCols = "Week")
dat2
# prodID storeID Term Exit Week_1 Week_2 Week_3 Week_4 Week_5
# 1: 1 1001 5 0 0 0 0 0 1
# 2: 1 1002 4 1 0 0 0 1 NA
# 3: 1 1003 3 1 0 0 1 NA NA
# 4: 1 1004 5 0 0 0 0 0 1
# 5: 2 1001 4 1 0 0 0 1 NA
# 6: 2 1002 3 1 0 0 1 NA NA
# 7: 2 1003 5 0 0 0 0 0 1
# 8: 3 1001 4 1 0 0 0 1 NA
# 9: 3 1002 3 1 0 0 1 NA NA
# 10: 3 1003 5 0 0 0 0 0 1
# 11: 4 1001 4 1 0 0 0 1 NA
# 12: 4 1002 3 1 0 0 1 NA NA
# 13: 5 1001 5 0 0 0 0 0 1
# 14: 5 1002 4 1 0 0 0 1 NA
# 15: 5 1003 3 1 0 0 1 NA NA
Or use this tidyverse method. 或使用此tidyverse方法。 I like this one better than my previous one because this method does not require manually typing the week values. 我比上一个更好，因为这种方法不需要手动输入星期值。
library(dplyr)
library(tidyr)
library(purrr)
dat2 <- dat %>%
mutate(Week = map2(1, Term, `:`)) %>%
unnest() %>%
group_by(prodID, Term) %>%
mutate(Week_Value = as.integer(Week == max(Week)),
Week = paste0("Week", Week)) %>%
spread(Week, Week_Value) %>%
ungroup()
dat2
# # A tibble: 15 x 9
# prodID storeID Term Exit Week1 Week2 Week3 Week4 Week5
# <int> <int> <int> <int> <int> <int> <int> <int> <int>
# 1 1 1001 5 0 0 0 0 0 1
# 2 1 1002 4 1 0 0 0 1 NA
# 3 1 1003 3 1 0 0 1 NA NA
# 4 1 1004 5 0 0 0 0 0 1
# 5 2 1001 4 1 0 0 0 1 NA
# 6 2 1002 3 1 0 0 1 NA NA
# 7 2 1003 5 0 0 0 0 0 1
# 8 3 1001 4 1 0 0 0 1 NA
# 9 3 1002 3 1 0 0 1 NA NA
# 10 3 1003 5 0 0 0 0 0 1
# 11 4 1001 4 1 0 0 0 1 NA
# 12 4 1002 3 1 0 0 1 NA NA
# 13 5 1001 5 0 0 0 0 0 1
# 14 5 1002 4 1 0 0 0 1 NA
# 15 5 1003 3 1 0 0 1 NA NA
UPDATE 更新
We can use str_pad from the stringr package to pad 0 before spread the week column to sort the column name. 我们可以使用str_pad包中的stringr到pad 0，然后再扩展week列以对列名进行排序。
library(tidyverse)
dat2 <- dat %>%
mutate(Week = map2(1, Term, `:`)) %>%
unnest() %>%
group_by(prodID, Term) %>%
mutate(Week_Value = as.integer(Week == max(Week)),
Week = paste0("Week", str_pad(Week, width = 3, pad = "0"))) %>%
spread(Week, Week_Value) %>%
ungroup()
dat2
# # A tibble: 15 x 9
# prodID storeID Term Exit Week001 Week002 Week003 Week004 Week005
# <int> <int> <int> <int> <int> <int> <int> <int> <int>
# 1 1 1001 5 0 0 0 0 0 1
# 2 1 1002 4 1 0 0 0 1 NA
# 3 1 1003 3 1 0 0 1 NA NA
# 4 1 1004 5 0 0 0 0 0 1
# 5 2 1001 4 1 0 0 0 1 NA
# 6 2 1002 3 1 0 0 1 NA NA
# 7 2 1003 5 0 0 0 0 0 1
# 8 3 1001 4 1 0 0 0 1 NA
# 9 3 1002 3 1 0 0 1 NA NA
# 10 3 1003 5 0 0 0 0 0 1
# 11 4 1001 4 1 0 0 0 1 NA
# 12 4 1002 3 1 0 0 1 NA NA
# 13 5 1001 5 0 0 0 0 0 1
# 14 5 1002 4 1 0 0 0 1 NA
# 15 5 1003 3 1 0 0 1 NA NA
DATA 数据
dat <- read.table(text = "prodID storeID Term Exit
1 1001 5 0
1 1002 4 1
1 1003 3 1
1 1004 5 0
2 1001 4 1
2 1002 3 1
2 1003 5 0
3 1001 4 1
3 1002 3 1
3 1003 5 0
4 1001 4 1
4 1002 3 1
5 1001 5 0
5 1002 4 1
5 1003 3 1",
header = TRUE)
解决方案2
2 2018-06-25 18:27:35
Here is one option with base R where we loop through the 'Term', tabulate to get a 0s and 1 for each element, append NA at the end with length<- and rbind the list elements to create the columns of interest 这是一个带有base R选项，我们在其中循环遍历'Term'， tabulate以获取每个元素的0和1，在NA的末尾添加length<-并rbind list元素以创建感兴趣的列
dat[paste0("Week", 1:5)] <- do.call(rbind, lapply(dat$Term,
function(x) `length<-`(tabulate(x), max(dat$Term))))
dat
# prodID storeID Term Exit Week1 Week2 Week3 Week4 Week5
#1 1 1001 5 0 0 0 0 0 1
#2 1 1002 4 1 0 0 0 1 NA
#3 1 1003 3 1 0 0 1 NA NA
#4 1 1004 5 0 0 0 0 0 1
#5 2 1001 4 1 0 0 0 1 NA
#6 2 1002 3 1 0 0 1 NA NA
#7 2 1003 5 0 0 0 0 0 1
#8 3 1001 4 1 0 0 0 1 NA
#9 3 1002 3 1 0 0 1 NA NA
#10 3 1003 5 0 0 0 0 0 1
#11 4 1001 4 1 0 0 0 1 NA
#12 4 1002 3 1 0 0 1 NA NA
#13 5 1001 5 0 0 0 0 0 1
#14 5 1002 4 1 0 0 0 1 NA
#15 5 1003 3 1 0 0 1 NA NA
Or using the similar approach with tidyverse 或与tidyverse一起使用类似的方法
library(tidyverse)
dat %>%
mutate(Week = map(Term, ~
tabulate(.x) %>%
as.list %>%
set_names(paste0("Week", seq_along(.))) %>%
as_tibble)) %>%
unnest
# prodID storeID Term Exit Week1 Week2 Week3 Week4 Week5
#1 1 1001 5 0 0 0 0 0 1
#2 1 1002 4 1 0 0 0 1 NA
#3 1 1003 3 1 0 0 1 NA NA
#4 1 1004 5 0 0 0 0 0 1
#5 2 1001 4 1 0 0 0 1 NA
#6 2 1002 3 1 0 0 1 NA NA
#7 2 1003 5 0 0 0 0 0 1
#8 3 1001 4 1 0 0 0 1 NA
#9 3 1002 3 1 0 0 1 NA NA
#10 3 1003 5 0 0 0 0 0 1
#11 4 1001 4 1 0 0 0 1 NA
#12 4 1002 3 1 0 0 1 NA NA
#13 5 1001 5 0 0 0 0 0 1
#14 5 1002 4 1 0 0 0 1 NA
#15 5 1003 3 1 0 0 1 NA NA
解决方案3
2 2018-06-25 18:31:20
An option using dplyr::mutate_at and case_when can be based on finding subscript integer in column name using quo_name(quo(.)) and then checking if column number is more/equal/less than value of Term . 使用dplyr::mutate_at和case_when的选项可以基于使用quo_name(quo(.))在column name查找下标整数，然后检查列号是否大于/等于/小于Term值。
# First add additional columns based on maximum value of Term
df[,paste("Week", 1:max(df$Term), sep="")] <- NA
library(dplyr)
df %>% mutate_at(vars(starts_with("Week")), funs(case_when(
as.integer(sub(".*(\\d+)","\\1",quo_name(quo(.)))) < Term ~ 0L,
as.integer(sub(".*(\\d+)","\\1",quo_name(quo(.)))) == Term ~ 1L,
TRUE ~ NA_integer_
)))
# prodID storeID Term Exit Week1 Week2 Week3 Week4 Week5
# 1 1 1001 5 0 0 0 0 0 1
# 2 1 1002 4 1 0 0 0 1 NA
# 3 1 1003 3 1 0 0 1 NA NA
# 4 1 1004 5 0 0 0 0 0 1
# 5 2 1001 4 1 0 0 0 1 NA
# 6 2 1002 3 1 0 0 1 NA NA
# 7 2 1003 5 0 0 0 0 0 1
# 8 3 1001 4 1 0 0 0 1 NA
# 9 3 1002 3 1 0 0 1 NA NA
# 10 3 1003 5 0 0 0 0 0 1
# 11 4 1001 4 1 0 0 0 1 NA
# 12 4 1002 3 1 0 0 1 NA NA
# 13 5 1001 5 0 0 0 0 0 1
# 14 5 1002 4 1 0 0 0 1 NA
# 15 5 1003 3 1 0 0 1 NA NA
Data: 数据：
df <- read.table(text="
prodID storeID Term Exit
1 1001 5 0
1 1002 4 1
1 1003 3 1
1 1004 5 0
2 1001 4 1
2 1002 3 1
2 1003 5 0
3 1001 4 1
3 1002 3 1
3 1003 5 0
4 1001 4 1
4 1002 3 1
5 1001 5 0
5 1002 4 1
5 1003 3 1",
header = TRUE, stringsAsFactors = FALSE)
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