R sapply循环替换for循环 [英] R sapply loop to replace for loop

问题描述

之前，我已经成功地将 for循环切换为 sapply循环， 而且我知道(system.time())它们更快.

I have successfully switched for loops to sapply loops before, and I know for a fact (system.time()) that they are faster.

但是我的思想仍然以for循环的方式工作...

BUT my mind still works in a for loop way...

请帮助我将其转换为for循环大小写:

Please help me to convert this for loop case:

names.list <- c("Anna", "Ana", "Albert", "Albort", "Rob", "Robb", "Tommy", "Tommie")
misspell.list <- c("Anna", "Albort", "Robb", "Tommie")
fix.list <- c("Ana", "Albert", "Rob", "Tommy")

for(i in 1:length(fix.list)) {
        names.list[which(names.list == misspell.list[i])] <- fix.list[i]

}

names.list

到sapply()

到目前为止，我得到了:

So far, I got:

sapply(seq_along(fix.list), function(x)
        names.list[which(names.list == misspell.list[x])]  <- fix.list[x]
)

但是它只会返回原始向量.

But it only returns me the original vector.

谢谢！

misspell.list和fix.list由adist()波纹管自动创建，原始名称.list具有665个元素.我的for()解决方案返回length(unique(names.list)) = 653个元素

the misspell.list and fix.list were created automatically by adist() bellow and the original names.list has 665 elements. My for() solution returns length(unique(names.list)) = 653 elements

# will do another sapply() substitution here soon
for(i in 1:(length(names.list)-1)) {
        distancias[i] <- adist(names.list[i], names.list[i+1])
}

# fix list
misspell.list <- names.list[which(distancias < 2)]
fix.list <- names.list[which(distancias < 2) +1]

谢谢您，现在我是 sapply霸主，我在这里只是为了展示与adist()一起使用的其他 for-sapply 替代方法

EDIT 2: thanks to you, now I'm a sapply overlord and I'm here just to show my other for-sapply substitution used with adist()

nomes <- sort(unique(names.list))
distancias <- rep(10, length(nomes))

#adist() for finding misspelling
sapply(seq_along(nomes), 
       function(x) {
                if(x<length(nomes)) {
                        distancias[x] <<- adist(nomes[x], nomes[x+1])
                        }
        }
       )
# fix list
misspell.list <- names.list[which(distancias < 2)]
fix.list <- names.list[which(distancias < 2) +1]

您已经知道的另一部分，再次感谢！

The other part you already know, thanks again!

推荐答案

使用match的解决方案要好得多，但是就您要尝试执行的操作而言，这将起作用.首先，您不需要which.您还需要使用<<-运算符告诉循环中定义的内部函数使用全局环境，而不是其自己的局部环境-否则，它不会更改names.list，而只会更改其副本.

The solution using match is much better, but in terms of what you were trying to do, this will work. Firstly, you don't need the which. You also need to use the <<- operator to tell the internal function defined within the loop to use the global environment rather than its own local one - otherwise it does not change names.list, only its copy.

sapply(seq_along(fix.list), function(x)
  names.list[names.list == misspell.list[x]]  <<- fix.list[x]
)

names.list
[1] "Ana"    "Ana"    "Albert" "Albert" "Rob"    "Rob"    "Tommy"  "Tommy" 

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