本文介绍了在R中将一列拆分为2的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
I have this dataframe
CC.Number Date Time Accident.Type Location.1
1 12T008826 07/01/2012 1630 PD (39.26699, -76.560642)
2 12L005385 07/02/2012 1229 PD (39.000549, -76.399312)
3 12L005388 07/02/2012 1229 PD (39.00058, -76.399267)
4 12T008851 07/02/2012 445 PI (39.26367, -76.56648)
5 12T008858 07/02/2012 802 PD (39.240862, -76.599017)
6 12T008860 07/02/2012 832 PD (39.27022, -76.63926)
我想将Location.1列拆分为"alt"和"lng"列
I want to split the column Location.1 to "alt" and "lng" columns to be like
CC.Number Date Time Accident.Type alt lng
1 12T008826 07/01/2012 1630 PD 39.26699 -76.560642
2 12L005385 07/02/2012 1229 PD 39.000549 -76.399312
3 12L005388 07/02/2012 1229 PD 39.00058 -76.399267
我尝试了
location <- md$Location.1
location1 <- substring(location, 2)
location2 <- substr(location1, 1, nchar(location1)-1 )
location3 <- strsplit(location2, ",")
但停留在将location3从列表转换为数据框的过程中
but stuck at converting location3 from list to dataframe
我尝试了
ocdf<-data.frame(location2)
colnames(locdf)[1] = c("x")
df <- separate(location, col=x,into = c("lat","log"), sep = ",")
但是我得到一个错误
推荐答案
与也可以 tidyr 分开
separate from tidyr also works
library(tidyr)
# Sub out the parentheses
df$Location.1 <- gsub("[()]", "", df$Location.1)
separate(df, col = Location.1, into = c("lat","long"), sep = ",")
# CC.Number Date Time Accident.Type lat long
#1 12T008826 07/01/2012 1630 PD 39.26699 -76.560642
#2 12L005385 07/02/2012 1229 PD 39.000549 -76.399312
#3 12L005388 07/02/2012 1229 PD 39.00058 -76.399267
#4 12T008851 07/02/2012 445 PI 39.26367 -76.56648
#5 12T008858 07/02/2012 802 PD 39.240862 -76.599017
#6 12T008860 07/02/2012 832 PD 39.27022 -76.63926
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