Question

我在R中创建多步函数时遇到问题。当我在函数外使用它们并将代码插入到代码行中时，我的代码行会起作用，但是当我使用function（）并将变量插入到代码行，它通常是错误{'New York (NY)', 'United States (US)'}

例如，这是我现在正在使用的功能：

Error: Result must have length 370, not 0

我创建的所有函数都一遍又一遍地遇到这个问题。有什么建议吗？

best <- function(state, outcome) {
    stateuse <- hospitaloutcome %>% filter(State == state)
    stateusecols <- stateuse[,c(2,7,11,17,23)]
    stateusecols <- stateusecols %>% rename('heart attack' = 
                                                'Hospital.30.Day.Death..Mortality..Rates.from.Heart.Attack')
    stateusecols <- stateusecols %>% rename('heart failure' = 
                                                'Hospital.30.Day.Death..Mortality..Rates.from.Heart.Failure')
    stateusecols <- stateusecols %>% rename('pneumonia' = 
                                                'Hospital.30.Day.Death..Mortality..Rates.from.Pneumonia')
    stateusecols$'heart failure' <- as.numeric(stateusecols$'heart failure')
    stateusecols$'pneumonia' <- as.numeric(stateusecols$'pneumonia')
    value <- min(stateusecols$outcome , na.rm=TRUE)
    neededrow <- stateusecols %>% filter(stateusecols$outcome == value)
    hospital <- neededrow[,1]
    hospital
}

#But when I call best('TX','heart attack') it errors with the result must have length 370, not 0
#However when I just do this:

stateuse <- hospitaloutcome %>% filter(State == 'TX')
    stateusecols <- stateuse[,c(2,7,11,17,23)]
    stateusecols <- stateusecols %>% rename('heart attack' = 
                                                'Hospital.30.Day.Death..Mortality..Rates.from.Heart.Attack')
    stateusecols <- stateusecols %>% rename('heart failure' = 
                                                'Hospital.30.Day.Death..Mortality..Rates.from.Heart.Failure')
    stateusecols <- stateusecols %>% rename('pneumonia' = 
                                                'Hospital.30.Day.Death..Mortality..Rates.from.Pneumonia')
    stateusecols$'heart failure' <- as.numeric(stateusecols$'heart failure')
    stateusecols$'pneumonia' <- as.numeric(stateusecols$'pneumonia')
    value <- min(stateusecols$'heart attack' , na.rm=TRUE)
    neededrow <- stateusecols %>% filter(stateusecols$'heart attack' == value)
    hospital <- neededrow[,1]
    hospital
#I get my answer, same lines of code, without the function() and the variables I need are in the same 
#spots as the variables defined.

Answer 1

@ user2554330的解决方案

通过将$outcome的所有[,outcome,drop=TRUE]换出，该功能现在可以使用了。

谢谢！

在R中创建函数的问题

1 个答案: