Question

我有一个由data.frames组成的列表：

> list1
$HSP90AB1
         lineNum count
HSP90AB1       1     1
INMT           2     0
CKB            3     0
NR2E1          4     1
ME3            5     0
FAM162A        6     0
KIRREL2        7     0

$INMT
         lineNum count
HSP90AB1       1     1
INMT           2     0
CKB            3     0
NR2E1          4     0
ME3            5     0
FAM162A        6     0
KIRREL2        7     1

$CKB
         lineNum count
HSP90AB1       1     1
INMT           2     0
CKB            3     0
NR2E1          4     0
ME3            5     1
FAM162A        6     0
KIRREL2        7     0

$NR2E1
         lineNum count
HSP90AB1       1     0
INMT           2     0
CKB            3     0
NR2E1          4     0
ME3            5     0
FAM162A        6     0
KIRREL2        7     0

$ME3
             lineNum count
HSP90AB1       1     0
INMT           2     1
CKB            3     0
NR2E1          4     0
ME3            5     0
FAM162A        6     0
KIRREL2        7     0

$FAM162A
         lineNum count
HSP90AB1       1     1
INMT           2     0
CKB            3     0
NR2E1          4     0
ME3            5     0
FAM162A        6     0
KIRREL2        7     0

$KIRREL2
         lineNum count
HSP90AB1       1     0
INMT           2     0
CKB            3     0
NR2E1          4     0
ME3            5     0
FAM162A        6     0
KIRREL2        7     0

我想获得以下结果：首先，将列表中每个元素的列1中的count的数量相加，然后将该次数的总和相加此元素的名称在列表的其余元素中有1，例如对于list1[[2]]（$ INMT），第一步将是：2（有两个1 in count列），第二步是：1（1中有一个与INMT对应的$ME3。列表的一个组成部分通过使用Reduce指出了我，但如果我这样做：

Reduce('+', list1)
             lineNum count
HSP90AB1       7     4
INMT          14     1
CKB           21     0
NR2E1         28     1
ME3           35     1
FAM162A       42     0
KIRREL2       49     1

我只是从我想要的第二步，任何想法？

由于

PS还有另一个需要解决的重要问题，在第一步中只应计算与名称不匹配的元素，仅在$HSP90AB1（list1[[1]]）的第一步中的示例中应计入与1对应的NR2E1，因为其他1对应HSP90AB1与list1[[1]]的名称匹配。

所需的输出如下：

HSP90AB1       4     
INMT           3     
CKB            2    
NR2E1          1     
ME3            2       
FAM162A        1      
KIRREL2        1

以下是dput(list1)：

structure(list(HSP90AB1 = structure(list(lineNum = 1:7, count = c(1L, 0L, 0L, 1L, 0L, 0L, 0L)),
.Names = c("lineNum", "count"), row.names = c("HSP90AB1","INMT", "CKB", "NR2E1", "ME3","FAM162A", "KIRREL2"), class = "data.frame"), 
INMT = structure(list(lineNum = 1:7, count = c(1L, 0L, 0L, 
0L, 0L, 0L, 1L)), .Names = c("lineNum", "count"), row.names = c("HSP90AB1", 
"INMT", "CKB", "NR2E1", "ME3", "FAM162A", "KIRREL2"), class = "data.frame"), 
CKB = structure(list(lineNum = 1:7, count = c(1L, 0L, 0L, 
0L, 1L, 0L, 0L)), .Names = c("lineNum", "count"), row.names = c("HSP90AB1", 
"INMT", "CKB", "NR2E1", "ME3", "FAM162A", "KIRREL2"), class = "data.frame"), 
NR2E1 = structure(list(lineNum = 1:7, count = c(0L, 0L, 0L, 
0L, 0L, 0L, 0L)), .Names = c("lineNum", "count"), row.names = c("HSP90AB1", 
"INMT", "CKB", "NR2E1", "ME3", "FAM162A", "KIRREL2"), class = "data.frame"), 
ME3 = structure(list(lineNum = 1:7, count = c(0L, 1L, 0L, 
0L, 0L, 0L, 0L)), .Names = c("lineNum", "count"), row.names = c("HSP90AB1", 
"INMT", "CKB", "NR2E1", "ME3", "FAM162A", "KIRREL2"), class = "data.frame"), 
FAM162A = structure(list(lineNum = 1:7, count = c(1L, 0L, 
0L, 0L, 0L, 0L, 0L)), .Names = c("lineNum", "count"), row.names = c("HSP90AB1", 
"INMT", "CKB", "NR2E1", "ME3", "FAM162A", "KIRREL2"), class = "data.frame"), 
KIRREL2 = structure(list(lineNum = 1:7, count = c(0L, 0L, 
0L, 0L, 0L, 0L, 0L)), .Names = c("lineNum", "count"), row.names = c("HSP90AB1", 
"INMT", "CKB", "NR2E1", "ME3", "FAM162A", "KIRREL2"), class = "data.frame")), .Names = c("HSP90AB1", "INMT", "CKB", "NR2E1", "ME3", "FAM162A", "KIRREL2"))

Answer 1

这应该这样做。一个重要的步骤是通过清空每个列表元素的不需要的项来修改列表：

list2 <- Map(function(x, exclude) {x[exclude, "count"] <- 0; x},
             list1, names(list1))

然后你可以这样做：

count1 <- sapply(list2, with, sum(count))
count2 <- Reduce(`+`, list2)$count

count1 + count2
# HSP90AB1     INMT      CKB    NR2E1      ME3  FAM162A  KIRREL2 
#        4        3        2        1        2        1        1

通过两个步骤计算列表中的元素

1 个答案: