R-如何相交（）并包含重复项？
我有以下字符字段，我正试图相交。这些应该是平等的R-如何相交（）并包含重复项？,r,character,intersect,R,Character,Intersect,我有以下字符字段，我正试图相交。这些应该是平等的 > char.y[[892]] [1] "E" "d" "w" "a" "r" "d" "s" " " "L" "i" "f" "e" "s" "c" "i" "e" "n" "c" "e" "s" > char.x[[892]] [1] "E" "d" "w" "a" "r" "d" "s" " " "L" "i" "f" "e" "s" "c" "i" "e" "n" "c" "e" "s" > intersect(ch
> char.y[[892]]
[1] "E" "d" "w" "a" "r" "d" "s" " " "L" "i" "f" "e" "s" "c" "i" "e" "n" "c" "e" "s"
> char.x[[892]]
[1] "E" "d" "w" "a" "r" "d" "s" " " "L" "i" "f" "e" "s" "c" "i" "e" "n" "c" "e" "s"
> intersect(char.x[[892]], char.y[[892]])
[1] "E" "d" "w" "a" "r" "s" " " "L" "i" "f" "e" "c" "n"
>
"E" "d" "w" "a" "r" "d" "s" " " "L" "i" "f" "e" "s" "c" "i" "e" "n" "c" "e"
它删除了我相信的两个条目，因此它已经返回了一个“s”，并且不会再执行第二次操作它删除了我相信的两个条目，因此它已经返回了一个“s”，并且不会再执行第二次操作使用
intersectsa <- c("E", "d", "w", "a", "r", "d", "s", " ", "L", "i", "f", "e", "s", "c", "i", "e", "n", "c", "e", "s")
b <- c("E", "d", "w", "a", "r", "d", "s", " ", "L", "i", "f", "e", "s", "c", "i", "e", "n", "c", "e", "s")
a[a %in% intersect(a, b)]
# [1] "E" "d" "w" "a" "r" "d" "s" " " "L" "i" "f" "e" "s" "c" "i" "e" "n" "c" "e" "s"
a使用intersect
将返回公共元素，但不会复制它们。例如，s
在其中3次，但只在相交处出现一次
例如，如果要查看删除了不相交值的同一布局，可以使用以下选项：
a <- c("E", "d", "w", "a", "r", "d", "s", " ", "L", "i", "f", "e", "s", "c", "i", "e", "n", "c", "e", "s")
b <- c("E", "d", "w", "a", "r", "d", "s", " ", "L", "i", "f", "e", "s", "c", "i", "e", "n", "c", "e", "s")
a[a %in% intersect(a, b)]
# [1] "E" "d" "w" "a" "r" "d" "s" " " "L" "i" "f" "e" "s" "c" "i" "e" "n" "c" "e" "s"
a这完全取决于所比较的向量（以及顺序），但这是否足够
b <- a <- c('E', 'd', 'w', 'a', 'r', 'd', 's', '', 'L', 'i', 'f', 'e', 's', 'c', 'i', 'e', 'n', 'c', 'e')
c <- letters[sample(1:26,100, rep=T)]
a[is.element(a,b)]
# [1] "E" "d" "w" "a" "r" "d" "s" "" "L" "i" "f" "e" "s" "c" "i" "e" "n" "c" "e"
a[is.element(a,c)]
# [1] "d" "w" "a" "r" "d" "s" "i" "f" "e" "s" "c" "i" "e" "n" "c" "e"
b这完全取决于您正在比较的向量（以及顺序），但这是否足够
b <- a <- c('E', 'd', 'w', 'a', 'r', 'd', 's', '', 'L', 'i', 'f', 'e', 's', 'c', 'i', 'e', 'n', 'c', 'e')
c <- letters[sample(1:26,100, rep=T)]
a[is.element(a,b)]
# [1] "E" "d" "w" "a" "r" "d" "s" "" "L" "i" "f" "e" "s" "c" "i" "e" "n" "c" "e"
a[is.element(a,c)]
# [1] "d" "w" "a" "r" "d" "s" "i" "f" "e" "s" "c" "i" "e" "n" "c" "e"
b我遇到了完全相同的问题，没有找到解决方案，因此我创建了自己的小函数“intersectdup”：
intersectdup我遇到了完全相同的问题，没有找到解决方案，因此我创建了自己的小函数“intersectdup”：
intersectdup了解克莱门斯，以下是基于c的结构中的一个简单函数：
intersectMe = function(x, y, duplicates=TRUE)
{
xyi = intersect(x,y);
if(!duplicates) { return (xyi); }
res = c();
for(xy in xyi)
{
y.xy = which(y == xy); ny.xy = length(y.xy);
x.xy = which(x == xy); nx.xy = length(x.xy);
min.xy = min(ny.xy, nx.xy);
res = c(res, rep(xy, min.xy) );
}
res;
}
说到克莱门斯，这里有一个基于c的结构中的简单函数：
intersectMe = function(x, y, duplicates=TRUE)
{
xyi = intersect(x,y);
if(!duplicates) { return (xyi); }
res = c();
for(xy in xyi)
{
y.xy = which(y == xy); ny.xy = length(y.xy);
x.xy = which(x == xy); nx.xy = length(x.xy);
min.xy = min(ny.xy, nx.xy);
res = c(res, rep(xy, min.xy) );
}
res;
}
vecsets
库也有帮助（使用Eric创建的示例）
vecsets
库也有帮助（使用Eric创建的示例）
也许您正在寻找setequal
<代码>相交
提供两个向量之间的公共（非重复）元素。例如，intersect（c（“a”，“b”），c（“b”，“c”）
给出了“b”
，但是输出结果应该是两个字符串中的公共字符。我需要的是这个答案，而不是一个正确/错误的答案，所以“不重复”是我的问题所在。有没有办法做到这一点并包括DUP？这是一个更复杂的问题，因为当向量是c（“b”，“b”）
和c（“b”）
时，你会怎么做？您希望返回的是“b”
还是c（“b”，“b”）
？可能您正在寻找setequal
<代码>相交
提供两个向量之间的公共（非重复）元素。例如，intersect（c（“a”，“b”），c（“b”，“c”）
给出了“b”
，但是输出结果应该是两个字符串中的公共字符。我需要的是这个答案，而不是一个正确/错误的答案，所以“不重复”是我的问题所在。有没有办法做到这一点并包括DUP？这是一个更复杂的问题，因为当向量是c（“b”，“b”）
和c（“b”）
时，你会怎么做？你想要返回的是“b”
还是c（“b”，“b”）
？vecsets:：vintersect（a，b）
别以为我以前用过vecsets
，谢谢你提醒我。返回的“E”“d”“d”“w”“a”“r”“s”“s”“L”“i”“f”“E”“E”“c”“c”“c”“n”
，这与OP预期的不同，我认为。vecsets:：vintersect（a，b）
不要认为我以前使用过vecsets
，谢谢你提醒我注意它。返回“E”“d”“d”“w”“a”“r”“s”“s”“L”“i”“f”“E”“E”“E”“c”“c”“c”“n”
，这与OP所期望的不同。您能说说为什么找不到解决方案，或者为什么以前接受的答案不能满足您的需要吗？GCF 126和108。。。获取主要因素并找到它们与重复项的交叉点。你说为什么你找不到解决方案，或者为什么以前接受的答案不符合你的需要？GCF为126和108。。。获取素数因子并找到它们与重复项的交集