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我有一个 json 文件，其行的格式如下：

{"checkin_info": {"11-3": 17, "8-5": 1, "15-0": 2, "15-3": 2, "15-5": 2, "14-4": 1, "14-    5": 3, "14-6": 6, "14-0": 2, "14-1": 2, "14-3": 2, "0-5": 1, "1-6": 1, "11-5": 3, "11-4": 11, "13-1": 1, "11-6": 6, "11-1": 18, "13-6": 5, "13-5": 4, "11-2": 9, "12-6": 5, "12-4": 8, "12-5": 5, "12-2": 12, "12-3": 19, "12-0": 20, "12-1": 14, "13-3": 1, "9-5": 2, "9-4": 1, "13-2": 6, "20-1": 1, "9-6": 4, "16-3": 1, "16-1": 1, "16-5": 1, "10-0": 3, "10-1": 4, "10-2": 4, "10-3": 4, "10-4": 1, "10-5": 2, "10-6": 2, "11-0": 3}, "type": "checkin", "business_id": "KO9CpaSPOoqm0iCWm5scmg"}

等等......它有8282个这样的条目。

我想把它转换成这样的csv文件。

  business_id              "0-0" "1-0" "2-0" "3-0" ….. "23-0" "0-1" ……. "23-1" …….. "0-4"    ……   "23-4" …… "23-6"

1 KO9CpaSPOoqm0iCWm5scmg     2    1     0     1  NA      1     1  NA     NA   NA    NA NA      6 NA      7

2 oRqBAYtcBYZHXA7G8FlPaA     1     2     2    NA  NA      2    NA  NA      1   NA     2 NA      2 NA      2

我试过这段代码：

urlc <- "C:\\Users\\Ayush\\Desktop\\yelp_training_set\\yelp_training_set_checkin.json"
conc = file(urlc, "r")
inputc <- readLines(conc, -1L)
usec <- lapply(X=inputc,fromJSON)
for (i in 1:8282)
   {
     tt<-usec[[i]]$checkin_info
     bb<-toString(tt)
     usec[[i]]$checkin_info<-bb
   }
dfc <- data.frame(matrix(unlist(usec), nrow=length(usec), byrow=T))
write.csv(dfc,file="checkin_tr.csv")

将其转换为如下形式：

                                                                                 X1                                                                      
business_id 
                                  1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1 

D0IB17N66FiyYDCzTlAI4A
                                                                      1, 1, 2, 1, 1     

HLQGo3EaYVvAv22bONGkIw
                                                                         1, 1, 1, 1 

J6OojF0R_1OuwNlrZI-ynQ      2, 1, 2, 1, 2, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 2, 1, 2

但我希望上面“X1”列中的条目在单独的列中，如第一个表所示。

我怎样才能做到这一点？请帮忙

4

1 回答 1

0

使用RJSONIO您可以执行以下操作：

library(RJSONIO)
tt <- fromJSON(tt)
data.frame(business_id =tt$business_id,
           do.call(rbind,list(tt$checkin_info)))


           business_id X11.3 X8.5 X15.0 X15.3 X15.5 X14.4 X14.5 X14.6 X14.0 X14.1 X14.3 X0.5 X1.6 X11.5 X11.4 X13.1 X11.6 X11.1 X13.6 X13.5 X11.2 X12.6 X12.4
1 KO9CpaSPOoqm0iCWm5scmg    17    1     2     2     2     1     3     6     2     2     2    1    1     3    11     1     6    18     5     4     9     5     8
  X12.5 X12.2 X12.3 X12.0 X12.1 X13.3 X9.5 X9.4 X13.2 X20.1 X9.6 X16.3 X16.1 X16.5 X10.0 X10.1 X10.2 X10.3 X10.4 X10.5 X10.6 X11.0
1     5    12    19    20    14     1    2    1     6     1    4     1     1     1     3     4     4     4     1     2     2     3

编辑

我在这里使用了一个新的想法。创建长格式 data.frame 然后使用reshape2例如将其转换为宽格式更容易。

library(RJSONIO)
## I create 2 shorter lines with different id
tt <-  '{"checkin_info": {"11-3": 17, "8-5": 1, "15-0": 2}, "type": "checkin", "business_id": "KO9CpaSPOoqm0iCWm5scmg"}'
tt1 <- '{"checkin_info": {"12-0": 17, "7-5": 1, "15-0": 5}, "type": "checkin", "business_id": "iddd2"}'
## use inputc <- readLines(conc, -1L) in your case
inputc <- list(tt,tt1)
usec <- lapply(X=inputc,function(x){
  tt <- fromJSON(x)
  data.frame(business_id =tt$business_id,
             names = names(tt$checkin_info),
             values =unlist(tt$checkin_info))
})
## create a long data frame
dat <- do.call(rbind,usec)     
## put in the wide format
library(reshape2)
dcast(business_id~names,data=dat)

             business_id 11-3 15-0 8-5 12-0 7-5
1 KO9CpaSPOoqm0iCWm5scmg   17    2   1   NA  NA
2                  iddd2   NA    5  NA   17   1
于 2013-06-18T12:40:08.863 回答