矢量版/矢量化等于R中的等于循环

I have a vector of values, call it X, and a data frame, call it dat.fram. I want to run something like "grep" or "which" to find all the indices of dat.fram[,3] which match each of the elements of X.

我有一个值向量，称之为X，数据帧，称之为dat.fram。我想运行类似“grep”或“which”的东西来查找与X的每个元素匹配的dat.fram [，3]的所有索引。

This is the very inefficient for loop I have below. Notice that there are many observations in X and each member of "match.ind" can have zero or more matches. Also, dat.fram has over 1 million observations. Is there any way to use a vector function in R to make this process more efficient?

这是我在下面循环的非常低效的循环。请注意，X中有许多观察值，“match.ind”的每个成员可以有零个或多个匹配。此外，dat.fram有超过100万次观测。有没有办法在R中使用向量函数来提高这个过程的效率？

Ultimately, I need a list since I will pass the list to another function that will retrieve the appropriate values from dat.fram .

最终，我需要一个列表，因为我将列表传递给另一个函数，该函数将从dat.fram中检索适当的值。

Code:

码：

match.ind=list()

for(i in 1:150000){
    match.ind[[i]]=which(dat.fram[,3]==X[i])
}

1 个解决方案

#1

UPDATE:

更新：

Ok, wow, I just found an awesome way of doing this... it's really slick. Wondering if it's useful in other contexts...?!

好的，哇，我刚刚找到了一个很棒的方法来做到这一点...它真的很光滑。想知道它是否在其他环境中有用......？！

### define v as a sample column of data - you should define v to be 
### the column in the data frame you mentioned (data.fram[,3]) 

v = sample(1:150000, 1500000, rep=TRUE)

### now here's the trick: concatenate the indices for each possible value of v,
### to form mybiglist - the rownames of mybiglist give you the possible values
### of v, and the values in mybiglist give you the index points

mybiglist = tapply(seq_along(v),v,c)

### now you just want the parts of this that intersect with X... again I'll
### generate a random X but use whatever X you need to

X = sample(1:200000, 150000)
mylist = mybiglist[which(names(mybiglist)%in%X)]

And that's it! As a check, let's look at the first 3 rows of mylist:

就是这样！作为检查，让我们看看前3行的mylist：

> mylist[1:3]

$`1`
[1]  401143  494448  703954  757808 1364904 1485811

$`2`
[1]  230769  332970  389601  582724  804046  997184 1080412 1169588 1310105

$`4`
[1]  149021  282361  289661  456147  774672  944760  969734 1043875 1226377

There's a gap at 3, as 3 doesn't appear in X (even though it occurs in v). And the numbers listed against 4 are the index points in v where 4 appears:

在3处有一个间隙，因为3中没有出现3（即使它出现在v中）。并且针对4列出的数字是v中的索引点，其中4出现：

> which(X==3)
integer(0)

> which(v==3)
[1]  102194  424873  468660  593570  713547  769309  786156  828021  870796  
883932 1036943 1246745 1381907 1437148

> which(v==4)
[1]  149021  282361  289661  456147  774672  944760  969734 1043875 1226377

Finally, it's worth noting that values that appear in X but not in v won't have an entry in the list, but this is presumably what you want anyway as they're NULL!

最后，值得注意的是，出现在X但不在v中的值在列表中没有条目，但这可能是你想要的，因为它们是NULL！

Extra note: You can use the code below to create an NA entry for each member of X not in v...

额外注意：您可以使用以下代码为X的每个成员创建一个NA条目，而不是v ...

blanks = sort(setdiff(X,names(mylist)))
mylist_extras = rep(list(NA),length(blanks))
names(mylist_extras) = blanks
mylist_all = c(mylist,mylist_extras)
mylist_all = mylist_all[order(as.numeric(names(mylist_all)))]

Fairly self-explanatory: mylist_extras is a list with all the additional list stuff you need (the names are the values of X not featuring in names(mylist), and the actual entries in the list are simply NA). The final two lines firstly merge mylist and mylist_extras, and then perform a reordering so that the names in mylist_all are in numeric order. These names should then match exactly the (unique) values in the vector X.

相当不言自明：mylist_extras是一个列表，其中包含您需要的所有其他列表内容（名称是名称中没有特征的X值（mylist），列表中的实际条目只是NA）。最后两行首先合并mylist和mylist_extras，然后执行重新排序，以便mylist_all中的名称按数字顺序排列。然后，这些名称应完全匹配向量X中的（唯一）值。

Cheers! :)

干杯! :)

ORIGINAL POST BELOW... superseded by the above, obviously!

ORIGINAL POST BELOW ...明显被上面取代了！

Here's a toy example with tapply that might well run significantly quicker... I made X and d relatively small so you could see what's going on:

这是一个tapply的玩具示例，可能会更快地运行...我使X和d相对较小，所以你可以看到发生了什么：

X = 3:7
n = 100
d = data.frame(a = sample(1:10,n,rep=TRUE), b = sample(1:10,n,rep=TRUE), 
               c = sample(1:10,n,rep=TRUE), stringsAsFactors = FALSE)

tapply(X,X,function(x) {which(d[,3]==x)})

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