在r中根据data.table中的多个条件创建一列[关闭]。

问题描述 投票:0回答:2

我希望使用data.table在其他两个列条件的基础上创建一个新的列。这是我的示例代码。

group <- c(1,1,1,2,2,2,3,3,3,4,4,4)  
date <- c(6,2,3,7,6,9,7,1,4,6,8,9)
val1<- c("","A","A","","A","A","","A","A","","A","A")

df1<-data.frame(group,date,val1)
dt1<-as.data.table(df1)

下面是输出结果

 group date val1
 1    6     
 1    2    A
 1    3    A
 2    7     
 2    6    A
 2    9    A
 3    7     
 3    1    A
 3    4    A
 4    6     
 4    8    A
 4    9    A

我想在每组(1,2,3,4)中找到日期的最小值 val1 = A,就像这样。

group date val1 findmin
 1    6             
 1    2    A       Y
 1    3    A        
 2    7             
 2    6    A       Y
 2    9    A        
 3    7             
 3    1    A       Y
 3    4    A        
 4    6             
 4    8    A       Y
 4    9    A        

我试过

dt1[,findmin:= ifelse(date=min(date[val1 == "A"])),"Y","", by = group]

读作:如果日期最小日期,其中val1="A",则在新的一列名为'findmin'的列中放一个 "Y",否则什么都不放,对每组(1,2,3,4)都这样做。我得到这个错误。

Error in `[.data.table`(dt1, , `:=`(findmin, ifelse(min(date[val1 == "A"]))),  : 
  Provide either by= or keyby= but not both

我很感激你的帮助,谢谢!

r if-statement data.table min
2个回答
1
投票

你必须注意你的括号,平等性要用 ==:

dt1[,findmin := fifelse(date == min(date[val1 == "A"]), "Y", ""), by = group]

1
投票

这段代码用dplyr就可以了。 我相信有一个更优雅的方法可以做到这一点。

if (!require(dplyr)) {
  install.packages("dplyr")
}
library(dplyr)
if (!require(data.table)) {
  install.packages("data.table")
}
library(data.table)

group <- c(1,1,1,2,2,2,3,3,3,4,4,4)  
date <- c(6,2,3,7,6,9,7,1,4,6,8,9)
val1<- c("","A","A","","A","A","","A","A","","A","A")

df1<-data.frame(group,date,val1)
dt1<-as.data.table(df1)

# filter for A
df2 <- df1 %>% filter(val1 == "A")

# group by group, arrange by date, get the 1st row, ungroup, add findmin = Y
df3 <- df2 %>% group_by(group) %>% arrange(date) %>% slice(1) %>% ungroup() %>% mutate(findmin = "Y", )

# join back to the original data
df4 <- df1 %>% left_join(df3, by = c("group", "date", "val1"))

# set NA in findmin to "" if you want
df5 <- df4 %>% mutate(findmin = ifelse(is.na(findmin), "", findmin))

# print
df5

   group date val1 findmin
1      1    6             
2      1    2    A       Y
3      1    3    A        
4      2    7             
5      2    6    A       Y
6      2    9    A        
7      3    7             
8      3    1    A       Y
9      3    4    A        
10     4    6             
11     4    8    A       Y
12     4    9    A        

用随机数据进行测试

# test randomized
df6 <- sample_frac(df1, size=1)
df6

   group date val1
1      3    4    A
2      3    1    A
3      4    8    A
4      4    6     
5      4    9    A
6      2    9    A
7      2    7     
8      3    7     
9      1    3    A
10     1    6     
11     2    6    A
12     1    2    A

df6 <- df6 %>% 
  filter(val1 == "A") %>% 
  group_by(group) %>%
  arrange(date) %>%
  slice(1) %>%
  ungroup() %>%
  mutate(findmin = "Y", )
df7 <- df1 %>%
  left_join(df6, by = c("group", "date", "val1")) %>%
  mutate(findmin = ifelse(is.na(findmin), "", findmin)) %>%
  arrange(group, val1, date, findmin)

df7

   group date val1 findmin
1      1    6             
2      1    2    A       Y
3      1    3    A        
4      2    7             
5      2    6    A       Y
6      2    9    A        
7      3    7             
8      3    1    A       Y
9      3    4    A        
10     4    6             
11     4    8    A       Y
12     4    9    A     

用which.min代替arrang and slice。

df6 <- sample_frac(df1, size=1)
df6
df6 <- df6 %>% 
  filter(val1 == "A") %>% 
  group_by(group) %>%
  slice(which.min(date)) %>% 
  ungroup() %>%
  mutate(findmin = "Y", )
df7 <- df1 %>%
  left_join(df6, by = c("group", "date", "val1")) %>%
  mutate(findmin = ifelse(is.na(findmin), "", findmin)) %>%
  arrange(group, val1, date, findmin)

df7

   group date val1 findmin
1      1    6             
2      1    2    A       Y
3      1    3    A        
4      2    7             
5      2    6    A       Y
6      2    9    A        
7      3    7             
8      3    1    A       Y
9      3    4    A        
10     4    6             
11     4    8    A       Y
12     4    9    A        
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