铸造一个数据帧

-1

structure(list(FROM_STOP_ID = c("DEP10001", "DEP10001", "DEP10001", 
          "DEP10001", "DEP10001", "DEP10001", 
      "DEP10001", "DEP10001", "DEP10001", 
          "DEP10001"), 
      TO_STOP_ID = c("DEP10001", "DEP10022", 
      "DEP25005", "DEP35024", "DEP45024", "DEP50002", 
      "STP62369", "STP62829", "STP63844", "STP63889"), 
      TRAVEL_TIME = c(0, 518, 497, 131, 0, 131, 483, 137, 
          489, 504)), .Names = c("FROM_STOP_ID",  "TO_STOP_ID", "TRAVEL_TIME"

预计outpout格式

从输入datframe为rownames，第二列的列名，第三列的值一栏填写了

structure(list(DEP10001 = 0, DEP10022 = 518, DEP25005 = 497, 
DEP35024 = 131, DEP45024 = 0), .Names = c("DEP10001", "DEP10022", 
"DEP25005", "DEP35024", "DEP45024"), row.names = "DEP10001", class = "data.frame")

来源

2016-10-11 Eswar

上面的第一个data.frame格式不好，但是'tidyr :: spread（df1，TO_STOP_ID，TRAVEL_TIME）'或者其他任何long到wide的变体。 – alistaire

我们可以用xtabs从base R

xtabs(TRAVEL_TIME~FROM_STOP_ID + TO_STOP_ID, df1N)

来源

2016-10-11 06:15:26 akrun

as.data.frame（acast（travel_time，FROM_STOP_ID_TO_STOP_ID，value.var =“TRAVEL_TIME”）） – Eswar

这对我更好。谢谢 – Eswar

@Eswar你必须加载一个库。 – akrun

铸造一个数据帧

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