1

次のJSONがあります

file.txt

[{"metric":"create","tags":{"host":"sdsn13","cluster":"cdw","type":"SG"},"aggregateTags":[],"dps" :{"1417621083":71.72777777777777,"1417621204":70.76859504132231,"1417621384":70.92222222222222,"1417621564":70.84444444444445,"1417621623":71.32203389830508,"1417621803":70.92777777777778,"1417621925":70.60655737704919,"1417621983":71.17241379310344, "1417622043":70.96666666666667,"1417622223":90.05555555555556,"1417622403":96.81666666666666,"1417622464":95.36065573770492,"1417622644":96.65,"1417622824":80.13333333333334,"1417623003":67.6536312849162,"1417623363":67.375,"1417623424":66.39344262295081,"1417623543":67.6890756302521,"1417623724":67.29834254143647,"1417623784":66.78333333333333,"1417623964":68.99444444444444,"1417624144":71.47777777777777,"1417624323":71.71508379888269,"1417624503":71.7611111111111,"1417624563" :66.66666666666667}}]

私はjsonliteを使用し、JSONを読み取ってデータフレームに変換することができました

k <- jsonlite::fromJSON("file.txt",simplifyDataFrame= T)
> head(k)
                            metric          tags.host tags.cluster     tags.type dps.1417621083 dps.1417621204 dps.1417621384 dps.1417621564
1 create sdsn13     cdw SG       71.72778        70.7686       70.92222       70.84444
  dps.1417621623 dps.1417621803 dps.1417621925 dps.1417621983 dps.1417622043 dps.1417622223 dps.1417622403 dps.1417622464 dps.1417622644
1       71.32203       70.92778       70.60656       71.17241       70.96667       90.05556       96.81667       95.36066          96.65
  dps.1417622824 dps.1417623003 dps.1417623363 dps.1417623424 dps.1417623543 dps.1417623724 dps.1417623784 dps.1417623964 dps.1417624144
1       80.13333       67.65363         67.375       66.39344       67.68908       67.29834       66.78333       68.99444       71.47778
  dps.1417624323 dps.1417624503 dps.1417624563
1       71.71508       71.76111       66.66667

私がやろうとしているのは、次の形式のデータ フレームを取得することです。

> head(k)
V1         V2        
1417621083 71.72778  
1417621204 70.76860  
1417621384 70.92222  
1417621564 70.84444

しかし、私は次のようになり、UNIXタイムスタンプを別のフィールドに分割する方法がわかりません.何らかの理由でそれが消えます:

try <- k[[3]]
try <- as.data.frame(t(try))
> head(try)
           1        
1417621083 71.72778  
1417621204 70.76860  
1417621384 70.92222  
1417621564 70.84444  

> colnames(try)
[1] "1"
colnames(try) <- c("one")

> data.frame(do.call('rbind', strsplit(as.character(try$one),' ',fixed=TRUE)))
   do.call..rbind...strsplit.as.character.try.one........fixed...TRUE..
1                                                      71.7277777777778
2                                                      70.7685950413223
3                                                      70.9222222222222
4                                                      70.8444444444444

library(stringr)
str_split_fixed(try$one, " ", 2)
      [,1]               [,2]
 [1,] "71.7277777777778" ""  
 [2,] "70.7685950413223" ""  
 [3,] "70.9222222222222" ""  
 [4,] "70.8444444444444" ""  
 [5,] "71.3220338983051" ""

ここで何か不足している場合は、助けてもらえますか?

4

1 に答える 1

0

あなたは試すことができます

 Unl <- unlist(k$dps)
 k1 <- data.frame(V1=names(Unl), V2=Unl, stringsAsFactors=FALSE)
 row.names(k1) <- NULL
 head(k1)
 #        V1       V2
 #1 1417621083 71.72778
 #2 1417621204 70.76860
 #3 1417621384 70.92222
 #4 1417621564 70.84444
 #5 1417621623 71.32203
 #6 1417621803 70.92778

 str(k1)
 #'data.frame': 25 obs. of  2 variables:
 #$ V1: chr  "1417621083" "1417621204" "1417621384" "1417621564" ...
 #$ V2: num  71.7 70.8 70.9 70.8 71.3 ...

アップデート

stackまたは、列名を使用して変更することもできます。

  stack(k$dps)
于 2014-12-04T17:39:24.850 に答える