Question

 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)

我想将列Location.1拆分为＆＃34; alt＆＃34;和＆＃34; lng＆＃34;列就像

  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

我试过

location <- md$Location.1
location1 <- substring(location, 2)
location2 <- substr(location1, 1, nchar(location1)-1 )
location3 <-  strsplit(location2, ",")

但坚持将location3从列表转换为dataframe

我试过

ocdf<-data.frame(location2)
colnames(locdf)[1] = c("x")
df <- separate(location, col=x,into = c("lat","log"), sep = ",")

但是我收到了错误

UseMethod出错（＆＃34;单独_＆＃34;）：没有适用的方法＆＃39;分离_＆＃39;应用于类＆＃34;字符＆＃34;
的对象

Answer 1

您也可以这样做，假设dat1是您的原始数据集名称，我们可以使用strsplit和gsub。首先，我们使用gsub替换逗号和括号，然后使用strsplit按空格分割值：

df1 <- setNames(data.frame(do.call("rbind",strsplit(gsub("\\(|\\)|,","",dat1$Location.1),split=" "))),c("Lat","Long"))
df2 <- data.frame(cbind(dat1[,1:(length(dat1)-1)],df1))

# CC.Number     Date Time Accident.Type       Lat       Long
# 1 12T008826 07/01/12 1630            PD  39.26699 -76.560642
# 2 12L005385 07/02/12 1229            PD 39.000549 -76.399312
# 3 12L005388 07/02/12 1229            PD  39.00058 -76.399267
# 4 12T008851 07/02/12  445            PI  39.26367  -76.56648
# 5 12T008858 07/02/12  802            PD 39.240862 -76.599017
# 6 12T008860 07/02/12  832            PD  39.27022  -76.63926

Answer 2

来自

handleError() {
  return Observable.of([false]);
}

的

separate也有效

tidyr

Answer 3

我们可以使用extract中的tidyr来捕获两个只包含带点数字元素的组，并丢弃“Location.1”中的其余元素

library(tidyr)
df1 %>% 
  extract(Location.1, into = c('alt', 'lng'), "\\(([0-9.]+),\\s+(-*[0-9.]+).")
# 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
#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

Answer 4

在 base 中，您可以使用trimws删除()，并使用read.table在,进行拆分。

cbind(md[1:4], read.table(sep=",", text=trimws(md$Location.1, whitespace = "[ ()]"),
 col.names=c("alt", "lng")))
#  CC.Number        Date Time  Accident.Type      alt       lng
#1 12T008826  07/01/2012 1630             PD 39.26699 -76.56064
#2 12L005385  07/02/2012 1229             PD 39.00055 -76.39931
#3 12L005388  07/02/2012 1229             PD 39.00058 -76.39927
#4 12T008851  07/02/2012  445             PI 39.26367 -76.56648
#5 12T008858  07/02/2012  802             PD 39.24086 -76.59902
#6 12T008860  07/02/2012  832             PD 39.27022 -76.63926

数据：

md <- structure(list(CC.Number = c("12T008826", "12L005385", "12L005388", 
"12T008851", "12T008858", "12T008860"), Date = c(" 07/01/2012", 
" 07/02/2012", " 07/02/2012", " 07/02/2012", " 07/02/2012", " 07/02/2012"
), Time = c(1630L, 1229L, 1229L, 445L, 802L, 832L), Accident.Type = c("            PD", 
"            PD", "            PD", "            PI", "            PD", 
"            PD"), Location.1 = c("  (39.26699, -76.560642)", 
" (39.000549, -76.399312)", "  (39.00058, -76.399267)", "   (39.26367, -76.56648)", 
" (39.240862, -76.599017)", "   (39.27022, -76.63926)")), class = "data.frame", row.names = c(NA, 
-6L))

在R中将列拆分为2

4 个答案: