1

我正在尝试遍历fevehicle()mpg 包中的函数，由：

https://github.com/rOpenGov/mpg

我一直在尝试为函数提供多个 vinid，甚至在循环之间给函数 5 秒的休息时间，以防万一，但我不断收到 HTTP 错误，即使单独运行，该函数也可以正常工作。任何想法可能是什么？下面是代码：

#using a loop
vin = c("19UUA86209A000532", "19UUA86239A021598", "19UUA8F20CA037748", "19UUA8F21CA008002", "19UUA8F21CA017878")
for (i in vin) {
  library(mpg)
  print(i)
  print(substr(i, 13, 17))
  q = substr(i, 13, 17)
  z = feVehicle(q)
  Sys.sleep(5)
  z = t(unlist(z))

}

or
#using lapply to see a difference
lapply(vin, feVehicle)

两者都抛出以下错误：

[1] "19UUA86209A000532"
[1] "00532"
failed to load HTTP resource
Error in t.default(unlist(z)) : argument is not a matrix
> lapply(vin, feVehicle)
failed to load HTTP resource
failed to load HTTP resource
failed to load HTTP resource
failed to load HTTP resource
failed to load HTTP resource

但是当我一次只运行一个时它工作正常：mpg::feVehicle(00532)

Vehicle data:
                                    value
atvType                            Diesel
barrels08              16.616739130434784
barrelsA08                            0.0
c240Dscr                             NULL
c240bDscr                            NULL
charge120                             0.0
charge240                             0.0
charge240b                            0.0
city08                                 21
city08U                               0.0
cityA08                                 0
cityA08U                              0.0
city
4

1 回答 1

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这是因为在您的单个示例中，您给出了一个数字，但在循环中您使用了一个字符：

#using a loop
vin = c("19UUA86209A000532", "19UUA86239A021598", "19UUA8F20CA037748", "19UUA8F21CA008002", "19UUA8F21CA017878")
for (i in vin) {
  library(mpg)
  print(i)
  print(substr(i, 13, 17))
  q = substr(i, 13, 17)
  z = feVehicle(as.numeric(q))
  Sys.sleep(5)
  z = t(unlist(z))

}

[1] "19UUA86209A000532"
[1] "00532"
[1] "19UUA86239A021598"
[1] "21598"
[1] "19UUA8F20CA037748"
[1] "37748"
[1] "19UUA8F21CA008002"
[1] "08002"
[1] "19UUA8F21CA017878"
[1] "17878"
于 2016-12-13T18:16:45.600 回答