Ruby 有类似 Python 的列表推导式吗? [英] Does Ruby have something like Python's list comprehensions?

问题描述

Python 有一个很好的特性:

Python has a nice feature:

print([j**2 for j in [2, 3, 4, 5]]) # => [4, 9, 16, 25]

在 Ruby 中它更简单:

In Ruby it's even simpler:

puts [2, 3, 4, 5].map{|j| j**2}

但如果是关于嵌套循环，Python 看起来更方便.

but if it's about nested loops Python looks more convenient.

在 Python 中我们可以这样做:

In Python we can do this:

digits = [1, 2, 3]
chars = ['a', 'b', 'c']    
print([str(d)+ch for d in digits for ch in chars if d >= 2 if ch == 'a'])    
# => ['2a', '3a']

Ruby 中的等价物是:

The equivalent in Ruby is:

digits = [1, 2, 3]
chars = ['a', 'b', 'c']
list = []
digits.each do |d|
    chars.each do |ch|
        list.push d.to_s << ch if d >= 2 && ch == 'a'
    end
end
puts list

Ruby 有类似的东西吗?

Does Ruby have something similar?

推荐答案

Ruby 中常用的方式是将 Enumerable 和 Array 方法实现相同:

The common way in Ruby is to properly combine Enumerable and Array methods to achieve the same:

digits.product(chars).select{ |d, ch| d >= 2 && ch == 'a' }.map(&:join)

这仅比列表推导式长 4 个左右的字符，并且同样具有表现力(恕我直言，当然，但由于列表推导式只是列表 monad 的一种特殊应用，有人可能会争辩说，使用Ruby 的集合方法)，同时不需要任何特殊的语法.

This is only 4 or so characters longer than the list comprehension and just as expressive (IMHO of course, but since list comprehensions are just a special application of the list monad, one could argue that it's probably possible to adequately rebuild that using Ruby's collection methods), while not needing any special syntax.

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