我有2个向量.
x=c("a", "b", "c", "d", "a", "b", "c")
y=structure(c(1, 2, 3, 4, 5, 6, 7, 8), .Names = c("a", "e", "b",
"c", "d", "a", "b", "c"))
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我想匹配a到a,b到b相应的序列,从而使x[2]匹配y[3],而不是y[7]; 和x[5]匹配y[6]而不是y[1]等等.
lapply(x, function(z) grep(z, names(y), fixed=T))
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得到:
[[1]]
[1] 1 6
[[2]]
[1] 3 7
[[3]]
[1] 4 8
[[4]]
[1] 5
[[5]]
[1] 1 6
[[6]]
[1] 3 7
[[7]]
[1] 4 …Run Code Online (Sandbox Code Playgroud) 匹配R中任何汉字的正则表达式是什么?
[\\p{Han}] 似乎没有按预期工作。
v=c("a","b","c","?","e","?")
grep("[\\p{Han}]",v, value = TRUE)
[1] "a"
Run Code Online (Sandbox Code Playgroud) 我们如何使用一个lambda或其他循环机制遍历一个列表以获得每个列表项的绝对值?
(defun span (start end &key (step 1))
(loop for n from start to end by step
collect n))
(setf bce #'(lambda (x) (abs x)) (span -10 -1))
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结果可以用作例如BCE时间轴。