免责声明 1：我不确定您的问题是关于如何计算每个子组的计数，还是如何绘制 5 组维恩图。我假设是后者。
免责声明 2：我发现 5 组维恩图非常难以阅读。到了无用的地步。但这是我个人的看法。
如果可以选择其他 R 包，这里有一个使用 VennDiagram 的 5 组示例（直接来自 VennDiagram reference manual）
library(VennDiagram);
venn.plot <- draw.quintuple.venn(
area1 = 301, area2 = 321, area3 = 311, area4 = 321, area5 = 301,
n12 = 188, n13 = 191, n14 = 184, n15 = 177,
n23 = 194, n24 = 197, n25 = 190,
n34 = 190, n35 = 173, n45 = 186,
n123 = 112, n124 = 108, n125 = 108,
n134 = 111, n135 = 104, n145 = 104,
n234 = 111, n235 = 107, n245 = 110,
n345 = 100,
n1234 = 61, n1235 = 60, n1245 = 59,
n1345 = 58, n2345 = 57,
n12345 = 31,
category = c("A", "B", "C", "D", "E"),
fill = c("dodgerblue", "goldenrod1", "darkorange1", "seagreen3", "orchid3"),
cat.col = c("dodgerblue", "goldenrod1", "darkorange1", "seagreen3", "orchid3"),
cat.cex = 2,
margin = 0.05,
cex = c(
1.5, 1.5, 1.5, 1.5, 1.5, 1, 0.8, 1, 0.8, 1, 0.8, 1, 0.8, 1, 0.8,
1, 0.55, 1, 0.55, 1, 0.55, 1, 0.55, 1, 0.55, 1, 1, 1, 1, 1, 1.5),
ind = TRUE);
png("venn_5set.png");
grid.draw(venn.plot);
dev.off();
更新 [2017 年 11 月 15 日]
您的源表采用非典型格式。正如我在我的 cmets 中解释的那样，您通常从 二进制矩阵（每组一列，每个观察的成员资格由 0 或 1 表示）或 集合元素列表。
老实说，我越来越不确定你到底想做什么。我有一种感觉，可能对维恩图有误解。例如，让我们看一下表格的第一行
# Read data
library(readxl);
data <- as.data.frame(read_excel("~/Downloads/dataset4venn.xlsx"));
rownames(data) <- data[, 1];
data <- data[, -1];
head(data);
# A B C D E
#1 8 8 7 8 10
#2 0 0 10 0 2
#3 0 0 0 0 3
#4 0 0 1 2 0
#5 1 0 1 0 2
#6 0 0 0 0 1
观察是特定组（即采样点）。 目击次数在这里并不重要：维恩图探索在不同地点采样的不同物种之间的逻辑关系，或者换句话说，哪些独特的物种是由站点 A-E 共享。
话虽如此，但忽略每个站点的目击次数，您可以在以下 5 组维恩图中显示不同站点之间的重叠。我首先定义了一个辅助函数cts 来计算每组/重叠的计数，然后将这些数字输入draw.quintuple.venn。
# Function to calculate the count per group/overlap
# Note: data is a global variable
cts <- function(set) {
df <- data;
for (i in 1:length(set)) df <- subset(df, df[set[i]] >= 1);
nrow(df);
}
# Plot
library(VennDiagram);
venn.plot <- draw.quintuple.venn(
area1 = cts("A"), area2 = cts("B"), area3 = cts("C"),
area4 = cts("D"), area5 = cts("E"),
n12 = cts(c("A", "B")), n13 = cts(c("A", "C")), n14 = cts(c("A", "D")),
n15 = cts(c("A", "E")), n23 = cts(c("B", "C")), n24 = cts(c("B", "D")),
n25 = cts(c("B", "E")), n34 = cts(c("C", "D")), n35 = cts(c("C", "E")),
n45 = cts(c("D", "E")),
n123 = cts(c("A", "B", "C")), n124 = cts(c("A", "B", "D")),
n125 = cts(c("A", "B", "E")), n134 = cts(c("A", "C", "D")),
n135 = cts(c("A", "C", "E")), n145 = cts(c("A", "D", "E")),
n234 = cts(c("B", "C", "D")), n235 = cts(c("B", "C", "E")),
n245 = cts(c("B", "D", "E")), n345 = cts(c("C", "D", "E")),
n1234 = cts(c("A", "B", "C", "D")), n1235 = cts(c("A", "B", "C", "E")),
n1245 = cts(c("A", "B", "D", "E")), n1345 = cts(c("A", "C", "D", "E")),
n2345 = cts(c("B", "C", "D", "E")),
n12345 = cts(c("A", "B", "C", "D", "E")),
category = c("A", "B", "C", "D", "E"),
fill = c("dodgerblue", "goldenrod1", "darkorange1", "seagreen3", "orchid3"),
cat.col = c("dodgerblue", "goldenrod1", "darkorange1", "seagreen3", "orchid3"),
cat.cex = 2,
margin = 0.05,
cex = c(
1.5, 1.5, 1.5, 1.5, 1.5, 1, 0.8, 1, 0.8, 1, 0.8, 1, 0.8, 1, 0.8,
1, 0.55, 1, 0.55, 1, 0.55, 1, 0.55, 1, 0.55, 1, 1, 1, 1, 1, 1.5),
ind = TRUE);
png("venn_5set.png");
grid.draw(venn.plot);
dev.off();
PS
各种 R 包/互联网源提供帮助函数来计算重叠，例如基于二进制矩阵或集合元素列表。例如，R/Bioconductor 包limma 提供了一个函数limma::vennCounts，它基于二进制矩阵计算所有重叠的计数。因此，如果您不想编写自己的函数（就像我一样），您也可以使用它们。无论哪种方式，对于更复杂的维恩图，我建议不要手动手动计算重叠，因为很容易出错（请参阅您的错误消息）。