如何在r中给定另一列的连续字符串的情况下找到列中连续数字的均值 [英] How to find the means of consecutive numbers in a column given consecutive string of another column in r

问题描述

我有一个与此类似的数据集:

I have a dataset that looks similar to this:

head(df,20)
   mmpd tot
1     0   0
2    mm   0
3    mm   1
4     0   0
5     0   0
6    mm   0
7    mm   1
8    mm   3
9    mm   1
10    0   0
11    0   0
12    0   0
13    0   0
14   mm   0
15   mm   0
16    0   0
17    0   0
18   mm   4
19   mm   1
20   mm   0

当df$mmpd 中的一串mm 对应时，我想得到df$tot 的平均值.因此，对于示例数据集，我想获得以下数字字符串:.5、1.25、0、1.667.df$mmpd 将始终是一串 mm > 1 或 0，并且该列可以以 0 或一串 mm 开头.

I would like to get the average of df$tot when it corresponds to a string of mm in df$mmpd. So for the example dataset, I'd like to get the following string of numbers: .5, 1.25, 0, 1.667. df$mmpd will always either be a string of mm > 1, or 0, and the column can begin with either 0 or a string of mm.

有没有办法在没有 for 循环的情况下做到这一点?

Is there a way to do this without a for loop?

推荐答案

Using data.table

library(data.table) # v 1.9.5+
setDT(df)[,.(my=mean(tot)), by=.(indx=rleid(mmpd),mmpd)][,indx:=NULL][mmpd=='mm']
   mmpd       my
#1:   mm 0.500000
#2:   mm 1.250000
#3:   mm 0.000000
#4:   mm 1.666667

显然，有很多方法可以做到(参见r 沿向量搜索并计算平均值).data.table 方法在此处速度最快且经过调整.

Apparently, there are many ways to do it (see r search along a vector and calculate the mean). The data.table method was fastest and adapted here.

注意:rleid 可以在 data.table 语法之外使用.这将更像传统"R 语法并产生相同的结果.

Note: rleid can be use outside of the data.table syntax. This will be more like "traditional" R syntax and produce the same results.

subset(aggregate(tot ~ indx + mmpd, 
          data=cbind(df,indx=rleid(df$mmpd)),
          FUN=mean),mmpd=="mm")

rleid(myrleid)不同生成方式的速度比较来自@JasonAizkalns 的回答).

Speed comparison of different ways to generate rleid (myrleid is from @JasonAizkalns answer).

> set.seed(1); x<-sample(1:2,100000,replace=T); 
  microbenchmark(rleid(x),
                 myrleid2=cumsum(c(1,diff(x)!=0)),
                 myrleid(x))
Unit: milliseconds
       expr      min       lq     mean   median       uq       max neval cld
   rleid(x) 1.422263 1.500873 1.586482 1.571315 1.662982  1.938254   100 a  
   myrleid2 3.860290 3.908308 4.369646 3.962497 4.177673 15.674611   100  b 
 myrleid(x) 7.282868 7.386515 7.753515 7.444008 7.654126 18.864898   100   c

对于非数字 x:

>  set.seed(1); x<-sample(c('a','b'),100000,replace=T); 
>  microbenchmark(rleid(x),myrleid2=cumsum(c(1,diff(as.numeric(factor(x)))!=0)),myrleid(x))
Unit: milliseconds
       expr       min        lq      mean    median       uq       max neval cld
   rleid(x)  1.465466  1.571662  1.684568  1.606614  1.66080  2.900983   100 a
   myrleid2  8.705447  9.276787 12.393393  9.907403 10.35032 61.080374   100  b
 myrleid(x) 11.970271 13.176144 18.779256 13.790767 14.09626 69.845587   100   c

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