添加列表中的值
Add up values in list
我目前正在处理一个包含以下数据的列表：
>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"
我想将每列的所有数字相加，只保留第 1 列中的值，该值出现在每一行中。
输出应如下所示：
>mergedData
[1] "x" "0" "9" "14" "13" "26"
我怎样才能做到这一点？
您可以使用这种方法：
c(resultsList[[1]][1],
colSums("mode<-"(do.call(rbind, resultsList)[ , -1], "numeric")))
# "x" "0" "9" "14" "13" "26"
这里，函数"mode<-"用于改变矩阵do.call(rbind, resultsList)[ , -1]的模式，包括用字符串表示的数字。
字符矩阵：
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"
数值矩阵：
"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) 类似于 mode(x) <- y 但不会更改 x 和 returns 结果。
编辑
这是一个解决方案，假设您所有的第 1 列字符串都采用 "var /// var2 /// ..." 形式。我们首先像这样恢复所有唯一变量：
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"
接下来，我们找出这些变量中的哪些包含在所有单独的行中：
matches <- sapply(1:length(uniqueVariables), function(x,y) grep(uniqueVariables[x],y), y=firstColumn)
variablesMatchingAllRows <- uniqueVariables[sapply(matches,"length")==length(resultsList)]
variablesMatchingAllRows
[1] "x"
然后我们将变量粘贴在一起（如果您有超过 1 个匹配所有行的变量）：
variablesMatchingAllRowsTest <- c("x","y","z")
paste(variablesMatchingAllRowsTest,collapse=" /// ")
[1] "x /// y /// z"
我们获得最终的第 1 列字符串并添加列总和：
> finalString <- paste(variablesMatchingAllRows,collapse=" /// ")
> c(finalString,colSums("mode<-"(do.call(rbind, resultsList)[ , -1], "numeric")))
[1] "x" "0" "9" "14" "13" "26"
旧答案
在下面的示例中，我们将首先在第 1 列中找到具有最小字符串大小的唯一字符串，然后我们将检查这个最小字符串是否包含在其他字符串中。然后我们将计算匹配行的列和。我们使用此数据作为示例：
> 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，我们找到 minimalString 和匹配此 minimalString 的相应行索引：
firstColumn <- sapply(resultsList,"[[",1)
minimalString <- unique(firstColumn[nchar(firstColumn)==min(nchar(firstColumn))])
indices <- grep(minimalString,firstColumn) # Grep on the first element in minimalString
我们得到：
> minimalString
[1] "x"
> indices
[1] 1 2 3 5
换句话说，除第 4 行外的所有行都与您的 minimalString 匹配。接下来我们在匹配的行上添加所有列和，如下所示：
> 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"
为了清楚起见，我们将进一步分解它：
内部 sapply(y,"[[",x)) 将获取列表 y 中索引 x 的所有元素，并将 return 它们作为向量。我们为 y = resultsList 和 x = 2:6 执行此操作。
请注意，我们还必须先将字符转换为数字：
> 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
接下来，我们计算与 indices:
匹配的行的列总和
> sums <- apply(intermediateResult[indices,],2,sum)
> sums
[1] 0 4 10 8 18
最后，我们还是要将和转换回字符，并在前面添加唯一的第1列标识符。我们得到：
> finalResult <- c(minimalString,as.character(sums))
> finalResult
[1] "x" "0" "4" "10" "8" "18"
对于您的示例，我们得到以下结果：
> 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"
我目前正在处理一个包含以下数据的列表：
>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"
我想将每列的所有数字相加，只保留第 1 列中的值，该值出现在每一行中。
输出应如下所示：
>mergedData
[1] "x" "0" "9" "14" "13" "26"
我怎样才能做到这一点？
您可以使用这种方法：
c(resultsList[[1]][1],
colSums("mode<-"(do.call(rbind, resultsList)[ , -1], "numeric")))
# "x" "0" "9" "14" "13" "26"
这里，函数"mode<-"用于改变矩阵do.call(rbind, resultsList)[ , -1]的模式，包括用字符串表示的数字。
字符矩阵：
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"
数值矩阵：
"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) 类似于 mode(x) <- y 但不会更改 x 和 returns 结果。
编辑 这是一个解决方案，假设您所有的第 1 列字符串都采用 "var /// var2 /// ..." 形式。我们首先像这样恢复所有唯一变量：
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"
接下来，我们找出这些变量中的哪些包含在所有单独的行中：
matches <- sapply(1:length(uniqueVariables), function(x,y) grep(uniqueVariables[x],y), y=firstColumn)
variablesMatchingAllRows <- uniqueVariables[sapply(matches,"length")==length(resultsList)]
variablesMatchingAllRows
[1] "x"
然后我们将变量粘贴在一起（如果您有超过 1 个匹配所有行的变量）：
variablesMatchingAllRowsTest <- c("x","y","z")
paste(variablesMatchingAllRowsTest,collapse=" /// ")
[1] "x /// y /// z"
我们获得最终的第 1 列字符串并添加列总和：
> finalString <- paste(variablesMatchingAllRows,collapse=" /// ")
> c(finalString,colSums("mode<-"(do.call(rbind, resultsList)[ , -1], "numeric")))
[1] "x" "0" "9" "14" "13" "26"
旧答案
在下面的示例中，我们将首先在第 1 列中找到具有最小字符串大小的唯一字符串，然后我们将检查这个最小字符串是否包含在其他字符串中。然后我们将计算匹配行的列和。我们使用此数据作为示例：
> 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，我们找到 minimalString 和匹配此 minimalString 的相应行索引：
firstColumn <- sapply(resultsList,"[[",1)
minimalString <- unique(firstColumn[nchar(firstColumn)==min(nchar(firstColumn))])
indices <- grep(minimalString,firstColumn) # Grep on the first element in minimalString
我们得到：
> minimalString
[1] "x"
> indices
[1] 1 2 3 5
换句话说，除第 4 行外的所有行都与您的 minimalString 匹配。接下来我们在匹配的行上添加所有列和，如下所示：
> 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"
为了清楚起见，我们将进一步分解它：
内部 sapply(y,"[[",x)) 将获取列表 y 中索引 x 的所有元素，并将 return 它们作为向量。我们为 y = resultsList 和 x = 2:6 执行此操作。
请注意，我们还必须先将字符转换为数字：
> 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
接下来，我们计算与 indices:
> sums <- apply(intermediateResult[indices,],2,sum)
> sums
[1] 0 4 10 8 18
最后，我们还是要将和转换回字符，并在前面添加唯一的第1列标识符。我们得到：
> finalResult <- c(minimalString,as.character(sums))
> finalResult
[1] "x" "0" "4" "10" "8" "18"
对于您的示例，我们得到以下结果：
> 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"