yon*_*ran 31 list-comprehension coffeescript
来自Python,我喜欢Coffeescript从Python和Perl借用的许多功能(范围/切片,理解,解构分配).Coffeescript中是否存在模仿Python enumerate
或zip
(itertools.izip
)函数的语法糖?
以下是我不太关心的模式:
# an enumerate call would be helpful here
i = 0
for x in arr
... use x and i ...
i++
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和
# a zip would be useful here
n = Math.min(arr1.length,arr2.length)
for i in 0...n
x = arr1[i]; y = arr2[i]
... use x and y ...
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Mar*_*ith 29
forEach有效内置:
a = ['a','b','c']
for el,i in a
alert "Element #{el} is at index #{i}"
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Jon*_*ier 13
枚举:
arr.forEach (x, i) ->
... use x and i ...
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zip
/ zipWith
(我从Haskell学到了这些;我认为它们在Python中的意思相同?):
zip = (arr1, arr2) ->
basic_zip = (el1, el2) -> [el1, el2]
zipWith basic_zip, arr1, arr2
zipWith = (func, arr1, arr2) ->
min = Math.min arr1.length, arr2.length
ret = []
for i in [0...min]
ret.push func(arr1[i], arr2[i])
ret
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一些例子(测试):
zip([1, 2, 3], [4, 5, 6]) # => [[1, 4], [2, 5], [3, 6]]
add = (el1, el2) -> el1 + el2
zipWith(add, [1, 2, 3], [4, 5, 6]) # => [5, 7, 9]
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更新:重新实现Haskell风格,只是为了好玩.没有模式匹配就不那么酷了,但是很好..
zipWith = (func, arr1, arr2) ->
return [] if arr1.length is 0 or arr2.length is 0
el1 = arr1.shift()
el2 = arr2.shift()
ret_arr = zipWith func, arr1, arr2
ret_arr.unshift func(el1, el2)
ret_arr
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哦,伙计,这很有趣.因此需要更多这样的问题:D
对于压缩和其他此类实用程序函数,Underscore.js几乎是标准库 - 它恰好是由CoffeeScript背后的人Jeremy Ashkenas创建的.有了它,您可以将您的zip示例编写为
for elems in _.zip(arr1, arr2)
x = elems[0]; y = elems[1]
...
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还是更好
for [x, y] in _.zip(arr1, arr2)
...
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使用模式匹配.但请注意,_.zip
使用的最大长度为arr1
而arr2
不是min; 因此,如果您不想处理undefined
值,则应首先截断较长的数组.
还有一个实施的CoffeeScript下划线的Underscore.coffee,这是看如果你想知道如何实现在CoffeeScript中特定回路的好地方.
CoffeeScript Cookbook列出了一个很好的zip函数实现:
# Usage: zip(arr1, arr2, arr3, ...)
zip = () ->
lengthArray = (arr.length for arr in arguments)
length = Math.min(lengthArray...)
for i in [0...length]
arr[i] for arr in arguments
zip([0, 1, 2, 3], [0, -1, -2, -3])
# => [[0, 0], [1, -1], [2, -2], [3, -3]]
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http://coffeescriptcookbook.com/chapters/arrays/zip-function