Question

我有一个如下所示的列表（这只是列表中的一个元素，每个州还有50个元素）

$AK
                                       Hospital.Name State   HeartAttack  HeartFailure     Pneumonia
99                  PROVIDENCE ALASKA MEDICAL CENTER    AK          13.4          12.4           7.9
103                         ALASKA REGIONAL HOSPITAL    AK          14.5          13.4           8.9
102                      FAIRBANKS MEMORIAL HOSPITAL    AK          15.5          15.6           9.8
106                     ALASKA NATIVE MEDICAL CENTER    AK          15.7          11.6          11.6
100                   MAT-SU REGIONAL MEDICAL CENTER    AK          17.7          11.4           9.0
104               YUKON KUSKOKWIM DELTA REG HOSPITAL    AK Not Available          11.2           6.9
110                         SITKA COMMUNITY HOSPITAL    AK Not Available Not Available           7.8
114 PEACEHEALTH KETCHIKAN MEDICAL             CENTER    AK Not Available          11.4           8.0
101                       BARTLETT REGIONAL HOSPITAL    AK Not Available          11.6           8.1
113                   NORTON SOUND REGIONAL HOSPITAL    AK Not Available Not Available           8.1
111             PROVIDENCE KODIAK ISLAND MEDICAL CTR    AK Not Available Not Available           8.2
115                         SOUTH PENINSULA HOSPITAL    AK Not Available          10.8           8.5
107                            MT EDGECUMBE HOSPITAL    AK Not Available Not Available           9.6
105               CENTRAL PENINSULA GENERAL HOSPITAL    AK Not Available          11.6           9.8
108                 PROVIDENCE VALDEZ MEDICAL CENTER    AK Not Available Not Available Not Available
109                       PROVIDENCE SEWARD HOSPITAL    AK Not Available Not Available Not Available
112                 CORDOVA COMMUNITY MEDICAL CENTER    AK Not Available Not Available Not Available

我正试图推出名单中HeartAttack率较低的医院。

所以当我编写一个函数来处理这个列表的第一个元素并调用

时

  test <- function(x){
    if(x[[1]][,3] == min(x[[1]][,3]))
      return(head(x[[1]][1],1))

  }

test(statedata)

似乎工作正常，但有以下警告

(Warning message:
In if (x[[1]][, 3] == min(x[[1]][, 3])) return(head(x[[1]][1], 1)) :
  the condition has length > 1 and only the first element will be used)

                       Hospital.Name
99 PROVIDENCE ALASKA MEDICAL CENTER

但是，当我尝试在lapply中使用相同的函数代码循环遍历整个列表元素时，它似乎无法正常工作并因以下问题而失败

result <- lapply(statedata,function(statedata) if(statedata[[1]][,3] == min(statedata[[1]][,3])) return(head(statedata[[1]][1],1)) )

Error in statedata[[1]][, 3] : incorrect number of dimensions

不确定我是否遗漏了lapply的内容？

Answer 1

这是正确的，因为你用函数参数声明数据掩盖了全局变量声明数据。更好的方法是：

statedata = lapply(statedata, function(s) {
 s[!is.na(as.integer(s$HeartAttack)),,drop=F]
})
wm = which.min(sapply(statedata, function(s) min(s$HeartAttack)))
wState = statedata[[wm]]
wState[which.min(wState$HeartAttack), "Hospital"]

第一个lapply是删除＆＃34; Not Available＆＃34;医院，接下来的事情就是找到最糟糕的状态，然后最坏的状态被用来抢到处境最差的最差的医院。

基于相同列表R的另一索引返回列表的特定索引

1 个答案: