基於R中其他列值和行數的平均列

Question

我正在使用R，並試圖根據列A中的值從另一個數據框中創建平均結果的新數據框。爲了演示我的目標，數據：

set.seed(1981) 
df <- data.frame(A = sample(c(0,1), replace=TRUE, size=100), 
B=round(runif(100), digits=4), 
C=sample(1:1000, 100, replace=TRUE)) 
head(df, 30) 

A B  C 
0 0.6739 459 
1 0.5466 178 
0 0.154 193 
0 0.41 206 
1 0.7526 791 
1 0.3104 679 
1 0.739 434 
1 0.421 171 
0 0.3653 577 
1 0.4035 739 
0 0.8796 147 
0 0.9138 37 
0 0.7257 350 
1 0.2125 779 
0 0.1502 495 
1 0.2972 504 
0 0.2406 245 
1 0.0325 613 
0 0.8642 539 
1 0.1096 630 
1 0.2113 363 
1 0.277 974 
0 0.0485 755 
1 0.0553 412 
0 0.509 24 
0 0.2934 795 
0 0.0725 413 
0 0.8723 606 
0 0.3192 591 
1 0.5557 177 

我需要通過計算B列和C列的平均值爲儘可能多的行作爲在列A的值以減少的數據的大小保持連續相同的，最多3個行。如果值A保持爲1，或者對於多於3行的值爲0，則它​​將滾動到新數據幀中的下一行，如下所示。

新的數據幀需要以下欄目：

Value of A B.Av   C.Av No. of rows used 
0   0.6739  459   1 
1   0.5466  178   1 
0   0.282  199.5   2 
1   0.600666667 634.6666667 3 
1   0.421  171   1 
0   0.3653  577   1 
1   0.4035  739   1 
0   0.8397  178   3 
1   0.2125  779   1 
0   0.1502  495   1 
1   0.2972  504   1 
0   0.2406  245   1 
1   0.0325  613   1 
0   0.8642  539   1 
1   0.1993  655.6666667 3 
0   0.0485  755   1 
1   0.0553  412   1 
0   0.291633333 410.6666667 3 
0   0.59575  598.5   2 
1   0.5557  177   1 

我沒有設法找到另一個類似的情形，以礦山，同時搜索堆棧溢出所以任何幫助將非常感激。

Answer 1

這裏是一個基礎-R解決方案：

## define a function to split the run-length if greater than 3 
split.3 <- function(l,v) { 
    o <- c(values=v,lengths=min(l,3)) 
    while(l > 3) { 
    l <- l - 3 
    o <- rbind(o,c(values=v,lengths=min(l,3))) 
    } 
    return(o) 
} 

## compute the run-length encoding of column A 
rl <- rle(df$A) 

## apply split.3 to the run-length encoding 
## the first column of vl are the values of column A 
## the second column of vl are the corresponding run-length limited to 3 
vl <- do.call(rbind,mapply(split.3,rl$lengths,rl$values)) 

## compute the begin and end row indices of df for each value of A to average 
fin <- cumsum(vl[,2]) 
beg <- fin - vl[,2] + 1 

## compute the averages 
out <- do.call(rbind,lapply(1:length(beg), function(i) data.frame(`Value of A`=vl[i,1], 
                    B.Av=mean(df$B[beg[i]:fin[i]]), 
                    C.Av=mean(df$C[beg[i]:fin[i]]), 
                    `No. of rows used`=fin[i]-beg[i]+1))) 
## Value.of.A  B.Av  C.Av No..of.rows.used 
##1   0 0.6739000 459.0000    1 
##2   1 0.5466000 178.0000    1 
##3   0 0.2820000 199.5000    2 
##4   1 0.6006667 634.6667    3 
##5   1 0.4210000 171.0000    1 
##6   0 0.3653000 577.0000    1 
##7   1 0.4035000 739.0000    1 
##8   0 0.8397000 178.0000    3 
##9   1 0.2125000 779.0000    1 
##10   0 0.1502000 495.0000    1 
##11   1 0.2972000 504.0000    1 
##12   0 0.2406000 245.0000    1 
##13   1 0.0325000 613.0000    1 
##14   0 0.8642000 539.0000    1 
##15   1 0.1993000 655.6667    3 
##16   0 0.0485000 755.0000    1 
##17   1 0.0553000 412.0000    1 
##18   0 0.2916333 410.6667    3 
##19   0 0.5957500 598.5000    2 
##20   1 0.5557000 177.0000    1 

Answer 2

這裏是一個data.table解決方案：

library(data.table) 
setDT(df) 
# create two group variables, consecutive A and for each consecutive A every three rows 
(df[,rleid := rleid(A)][, threeWindow := ((1:.N) - 1) %/% 3, rleid] 

# calculate the average of the columns grouped by the above two variables 
    [, c(.N, lapply(.SD, mean)), .(rleid, threeWindow)] 

# drop group variables 
    [, `:=`(rleid = NULL, threeWindow = NULL)][]) 

# N A   B  C 
#1: 1 0 0.6739000 459.0000 
#2: 1 1 0.5466000 178.0000 
#3: 2 0 0.2820000 199.5000 
#4: 3 1 0.6006667 634.6667 
#5: 1 1 0.4210000 171.0000 
#6: 1 0 0.3653000 577.0000 
#7: 1 1 0.4035000 739.0000 
#8: 3 0 0.8397000 178.0000 
#9: 1 1 0.2125000 779.0000 
#10: 1 0 0.1502000 495.0000 
#11: 1 1 0.2972000 504.0000 
#12: 1 0 0.2406000 245.0000 
#13: 1 1 0.0325000 613.0000 
#14: 1 0 0.8642000 539.0000 
#15: 3 1 0.1993000 655.6667 
#16: 1 0 0.0485000 755.0000 
#17: 1 1 0.0553000 412.0000 
#18: 3 0 0.2916333 410.6667 
#19: 2 0 0.5957500 598.5000 
#20: 1 1 0.5557000 177.0000