3

我认为这还没有被问到,但是有没有办法将具有多个级别和不均匀结构的列表信息组合成一个“长”格式的数据框?

具体来说:

library(XML)
library(plyr)
xml.inning <- "http://gd2.mlb.com/components/game/mlb/year_2009/month_05/day_02/gid_2009_05_02_chamlb_texmlb_1/inning/inning_5.xml"
xml.parse <- xmlInternalTreeParse(xml.inning)
xml.list <- xmlToList(xml.parse)
## $top$atbat
## $top$atbat$pitch
##             des              id            type               x               y 
##          "Ball"           "310"             "B"         "70.39"        "125.20" 

以下是结构:

> llply(xml.list, function(x) llply(x, function(x) table(names(x))))
$top
$top$atbat
.attrs  pitch 
     1      4 
$top$atbat
.attrs  pitch 
     1      4 
$top$atbat
.attrs  pitch 
     1      5 
$bottom
$bottom$action
     b    des  event      o  pitch player      s 
     1      1      1      1      1      1      1 
$bottom$atbat
.attrs  pitch 
     1      5 
$bottom$atbat
.attrs  pitch 
     1      5 
$bottom$atbat
.attrs  pitch runner 
     1      5      1 
$bottom$atbat
.attrs  pitch runner 
     1      7      1 
$.attrs
$.attrs$num
character(0)
$.attrs$away_team
character(0)
$.attrs$

我想要的是来自音高类别的命名向量的数据框,以及正确的(topatbatbottom)。因此,由于列数不同,我需要忽略不适合 data.frame 的级别。像这样的东西:

   first second third    des     x
1    top  atbat pitch   Ball 70.29
2    top  atbat pitch Strike 69.24
3 bottom  atbat pitch    Out 67.22

有没有一种优雅的方式来做到这一点?谢谢!

4

2 回答 2

5

我不知道优雅,但这有效。那些更熟悉 plyr 的人可能会提供更通用的解决方案。

cleanFun <- function(x) {
   a <- x[["atbat"]]
   b <- do.call(rbind,a[names(a)=="pitch"])
   c <- as.data.frame(b)
}
ldply(xml.list[c("top","bottom")], cleanFun)[,1:5]
     .id             des  id type      x
1    top            Ball 310    B  70.39
2    top   Called Strike 311    S 118.45
3    top   Called Strike 312    S  86.70
4    top In play, out(s) 313    X  79.83
5 bottom            Ball 335    B  15.45
6 bottom   Called Strike 336    S  77.25
7 bottom Swinging Strike 337    S  99.57
8 bottom            Ball 338    B 106.44
9 bottom In play, out(s) 339    X 134.76
于 2010-08-04T23:25:28.993 回答
1

.id功能ldply()很好,但是一旦你做另一个,它们似乎就会重叠ldply()

这是使用的相当通用的功能rbind.fill()

aho <- ldply(llply(xml.list[[1]], function(x) ldply(x, function(x) rbind.fill(data.frame(t(x))))))
> aho[1:5,1:4]
     .id                                                       des   id type
1  pitch                                                      Ball  310    B
2  pitch                                             Called Strike  311    S
3  pitch                                             Called Strike  312    S
4  pitch                                           In play, out(s)  313    X
5 .attrs Alexei Ramirez lines out to second baseman Ian Kinsler.   <NA> <NA>

缺少第二.idldply(),因为我们已经有一个.id. 我们可以通过将第.id一个命名为不同的名称来解决此问题,但它似乎不连贯。

aho2 <- ldply(llply(xml.list[[1]], function(x) {
  out <- ldply(x, function(x) rbind.fill(data.frame(t(x))))
  names(out)[1] <- ".id2"
  out
}))
> aho2[1:5,1:4]
    .id   .id2                                                       des   id
1 atbat  pitch                                                      Ball  310
2 atbat  pitch                                             Called Strike  311
3 atbat  pitch                                             Called Strike  312
4 atbat  pitch                                           In play, out(s)  313
5 atbat .attrs Alexei Ramirez lines out to second baseman Ian Kinsler.   <NA>
于 2010-08-05T19:00:09.460 回答