加起來值列表

Question

我目前正在與包括以下數據的列表的工作：

>resultsList 

$`1` 
[1] "x" "0"   "1"   "1"   "1"   "5"   
$`2` 
[1] "x /// y" "0"  "1"   "1"   "2"   "3"   
$`3` 
[1] "x" "0"   "1"   "3"   "2"   "4"   
$`4` 
[1] "x /// z" "0"  "1"   "2"   "2"   "2"   
$`5` 
[1] "x" "0"   "1"   "3"   "3"   "4"   
$`6` 
[1] "x" "0"   "0"   "0"   "1"   "2"   
$`7` 
[1] "x" "0"   "2"   "2"   "1"   "4"    
$`8` 
[1] "x /// y" "0"  "2"   "2"   "1"   "2" 

我想加起來每列的所有號碼，並只保留其中的第1列的值呈現在每一行中。

輸出應該是這樣的：

>mergedData 

[1] "x" "0" "9" "14" "13" "26" 

我怎樣才能做到這一點？

Answer 1

編輯 這是一個解決方案，它假定所有的第1列字符串都以「var /// var2 /// ...」的形式出現。我們首先恢復所有唯一變量是這樣的：

resultsList <- list(c("x","0","1","1","1","5"), 
        c("x /// y","0","1","1","2","3"), 
        c("x","0","1","3","2","4"), 
        c("x /// z","0","1","2","2","2"), 
        c("x","0","1","3","3","4"), 
        c("x","0","0","0","1","2"), 
        c("x","0","2","2","1","4"), 
        c("x /// y","0","2","2","1","2")) 


firstColumn <- sapply(resultsList,"[[",1) 
listsOfVariables <- c(strsplit(firstColumn," /// ")) 
vector <- c() 
for(i in 1:length(listsOfVariables)) 
{ 
    vector <- c(vector,listsOfVariables[[i]]) 
} 

uniqueVariables <- unique(vector) 
uniqueVariables 
[1] "x" "y" "z" 

接下來，我們找出這些變量都包含在所有各行的：

matches <- sapply(1:length(uniqueVariables), function(x,y) grep(uniqueVariables[x],y), y=firstColumn) 
variablesMatchingAllRows <- uniqueVariables[sapply(matches,"length")==length(resultsList)] 
variablesMatchingAllRows 
[1] "x" 

我們則變量粘貼在一起（如果你的所有行符合超過1個變量）：

variablesMatchingAllRowsTest <- c("x","y","z") 
paste(variablesMatchingAllRowsTest,collapse=" /// ") 
[1] "x /// y /// z" 

我們獲得最後一列1串並添加列總和：

> finalString <- paste(variablesMatchingAllRows,collapse=" /// ") 
> c(finalString,colSums("mode<-"(do.call(rbind, resultsList)[ , -1], "numeric"))) 
[1] "x" "0" "9" "14" "13" "26" 

OLD ANSWER

在下面的例子中，我們將首先發現在具有最小stringsize的第1列的唯一的字符串，然後我們將檢查這個最小的字符串包含在其他字符串。然後，我們將計算匹配行的列數。我們使用這個數據爲例：

> resultsList <- list(c("x","0","1","1","1","5"), 
+      c("a b x /// y","0","1","1","2","3"), 
+      c("x","0","1","3","2","4"), 
+      c("a /// z","0","1","3","3","4"), 
+      c("bd x","0","1","5","3","6")) 
> resultsList 
[[1]] 
[1] "x" "0" "1" "1" "1" "5" 

[[2]] 
[1] "a b x /// y" "0"   "1"   "1"   "2"   "3"   

[[3]] 
[1] "x" "0" "1" "3" "2" "4" 

[[4]] 
[1] "a /// z" "0"  "1"  "3"  "3"  "4"  

[[5]] 
[1] "bd x" "0" "1" "5" "3" "6" 

首先，我們找到了匹配這minimalString的minimalString和相應的行索引：

firstColumn <- sapply(resultsList,"[[",1) 
minimalString <- unique(firstColumn[nchar(firstColumn)==min(nchar(firstColumn))]) 
indices <- grep(minimalString,firstColumn) # Grep on the first element in minimalString 

我們得到：

> minimalString 
[1] "x" 
> indices 
[1] 1 2 3 5 

換句話說，除第4行外的所有行都與您的minimalString匹配。接下來，我們添加所有的columnsums在匹配的行這樣的：

> c(minimalString, as.character(apply(sapply(2:6,function(x,y,z) as.numeric(sapply(y,"[[",x)),y=resultsList)[indices,],2,sum))) 
[1] "x" "0" "4" "10" "8" "18" 

我們將進一步把它分解爲清楚：

內sapply(y,"[[",x))將獲取列表Y索引x的所有元素，並返回他們作爲一個載體。我們爲y = resultsList和x = 2:6這樣做。 請注意，我們也有人物首先轉換爲數學運算：

> intermediateResult <- sapply(2:6,function(x,y,z) as.numeric(sapply(y,"[[",x)),y=resultsList) 
> intermediateResult 
    [,1] [,2] [,3] [,4] [,5] 
[1,] 0 1 1 1 5 
[2,] 0 1 1 2 3 
[3,] 0 1 3 2 4 
[4,] 0 1 3 3 4 
[5,] 0 1 5 3 6 

接下來，我們計算出各行的匹配indices的columnsums：最後

> sums <- apply(intermediateResult[indices,],2,sum) 
> sums 
[1] 0 4 10 8 18 

，我們還是要轉換總結回字符並在前面添加唯一的第1列標識符。我們得到：

> finalResult <- c(minimalString,as.character(sums)) 
> finalResult 
[1] "x" "0" "4" "10" "8" "18" 

對於示例，我們得到如下結果：

> resultsList <- list(c("x","0","1","1","1","5"), 
+      c("x /// y","0","1","1","2","3"), 
+      c("x","0","1","3","2","4"), 
+      c("x /// z","0","1","2","2","2"), 
+      c("x","0","1","3","3","4"), 
+      c("x","0","0","0","1","2"), 
+      c("x","0","2","2","1","4"), 
+      c("x // y","0","2","2","1","2")) 
> firstColumn <- sapply(resultsList,"[[",1) 
> minimalString <- unique(firstColumn[nchar(firstColumn)==min(nchar(firstColumn))]) 
> indices <- grep(minimalString,firstColumn) # Grep on the first element in minimalString 
> minimalString 
[1] "x" 
> indices 
[1] 1 2 3 4 5 6 7 8 
> c(minimalString, as.character(apply(sapply(2:6,function(x,y,z) as.numeric(sapply(y,"[[",x)),y=resultsList)[indices,],2,sum))) 
[1] "x" "0" "9" "14" "13" "26" 

Answer 2

您可以使用此方法：

c(resultsList[[1]][1], 
    colSums("mode<-"(do.call(rbind, resultsList)[ , -1], "numeric"))) 
# "x" "0" "9" "14" "13" "26" 

---

這裏，功能"mode<-"是用來改變包括表示爲字符串的數字矩陣do.call(rbind, resultsList)[ , -1]的模式。

字符矩陣：

do.call(rbind, resultsList)[ , -1] 
#  [,1] [,2] [,3] [,4] [,5] 
# [1,] "0" "1" "1" "1" "5" 
# [2,] "0" "1" "1" "2" "3" 
# [3,] "0" "1" "3" "2" "4" 
# [4,] "0" "1" "2" "2" "2" 
# [5,] "0" "1" "3" "3" "4" 
# [6,] "0" "0" "0" "1" "2" 
# [7,] "0" "2" "2" "1" "4" 
# [8,] "0" "2" "2" "1" "2" 

的數字矩陣：

"mode<-"(do.call(rbind, resultsList)[ , -1], "numeric") 
#  [,1] [,2] [,3] [,4] [,5] 
# [1,] 0 1 1 1 5 
# [2,] 0 1 1 2 3 
# [3,] 0 1 3 2 4 
# [4,] 0 1 2 2 2 
# [5,] 0 1 3 3 4 
# [6,] 0 0 0 1 2 
# [7,] 0 2 2 1 4 
# [8,] 0 2 2 1 2 

命令"mode<-"(x, y)類似於mode(x) <- y但不改變x並返回結果。