将嵌套的Python循环转换为列表推导 [英] Transforming nested Python loops into list comprehensions

问题描述

我已经开始处理一些Euler项目的问题，并且已经解决了编号4 和一个简单的蛮力解决方案:

I've started working on some Project Euler problems, and have solved number 4 with a simple brute force solution:

def mprods(a,b):
 c = range(a,b)
 f = []
 for d in c:
  for e in c:
   f.append(d*e)
 return f

max([z for z in mprods(100,1000) if str(z)==(''.join([str(z)[-i] for i in range(1,len(str(z))+1)]))])

解决后，我试图使其尽可能紧凑，并提出了可怕的底线！

After solving, I tried to make it as compact as possible, and came up with that horrible bottom line!

不要遗漏一些东西，我正在尝试将mprods函数压缩为一个列表理解.到目前为止，我已经提出了以下尝试:

Not to leave something half-done, I am trying to condense the mprods function into a list comprehension. So far, I've come up with these attempts:

• [d*e for d,e in (range(a,b), range(a,b))]
  显然完全走在错误的轨道上. :-)
• [d*e for x in [e for e in range(1,5)] for d in range(1,5)]
  这给了我[4, 8, 12, 16, 4, 8, 12, 16, 4, 8, 12, 16, 4, 8, 12, 16]，在我期望的地方 [1, 2, 3, 4, 2, 4, 6, 8, 3, 6, 9, 12, 4, 8, 12, 16]或类似版本.

• [d*e for d,e in (range(a,b), range(a,b))]
  Obviously completely on the wrong track. :-)
• [d*e for x in [e for e in range(1,5)] for d in range(1,5)]
  This gives me [4, 8, 12, 16, 4, 8, 12, 16, 4, 8, 12, 16, 4, 8, 12, 16], where I expect [1, 2, 3, 4, 2, 4, 6, 8, 3, 6, 9, 12, 4, 8, 12, 16] or similar.

有没有Pythonistas可以帮助您? :)

Any Pythonistas out there that can help? :)

推荐答案

c = range(a, b)
print [d * e for d in c for e in c]

这篇关于将嵌套的Python循环转换为列表推导的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！