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
$KIRREL2
lineNum count
HSP90AB1 1 0
INMT 2 0
CKB 3 0
NR2E1 4 0
ME3 5 0
FAM162A 6 0
KIRREL2 7 0
我想获得以下结果：首先，对列表中每个元素1的列中count的总和进行累加，然后1对元素的其余元素中该元素的名称具有a的次数求和。例如list1[[2]]（$ INMT）的列表，第一步将是：2（1该count列中有两个），第二步将是：1（中有一个1对应INMT于$ME3）。列表的一部分通过使用指出了我的意思，Reduce但是如果我这样做：
Reduce('+', list1)
lineNum count
HSP90AB1 7 4
INMT 14 1
CKB 21 0
NR2E1 28 1
ME3 35 1
FAM162A 42 0
KIRREL2 49 1
我只从寻找的东西中得到第二步，有什么想法吗？
谢谢
PS还有一个重要的问题要解决，第一步只应计算与名称不匹配的元素，在我的示例中，第一步中仅对$HSP90AB1（list1[[1]]）进行1对应，NR2E1因为另一个1对应于HSP90AB1哪个匹配的名称list1[[1]]。
所需的输出如下：
HSP90AB1 4
INMT 3
CKB 2
NR2E1 1
ME3 2
FAM162A 1
KIRREL2 1
这是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"))
这应该做。重要的步骤是通过将每个列表元素的不需要项归零来修改列表：
list2 <- Map(function(x, exclude) {x[exclude, "count"] <- 0; x},
list1, names(list1))
然后，您可以执行以下操作：
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
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