91 lines
3.5 KiB
Python
91 lines
3.5 KiB
Python
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# _*_ coding: utf-8 _*_
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"""
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python_oneline.py by xianhu
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"""
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# 首先来个python之禅
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# python -c "import this"
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"""
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The Zen of Python, by Tim Peters
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Beautiful is better than ugly.
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Explicit is better than implicit.
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Simple is better than complex.
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Complex is better than complicated.
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Flat is better than nested.
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Sparse is better than dense.
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Readability counts.
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Special cases aren't special enough to break the rules.
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Although practicality beats purity.
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Errors should never pass silently.
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Unless explicitly silenced.
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In the face of ambiguity, refuse the temptation to guess.
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There should be one-- and preferably only one --obvious way to do it.
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Although that way may not be obvious at first unless you're Dutch.
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Now is better than never.
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Although never is often better than *right* now.
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If the implementation is hard to explain, it's a bad idea.
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If the implementation is easy to explain, it may be a good idea.
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Namespaces are one honking great idea -- let's do more of those!
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"""
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# 一行代码启动一个Web服务
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# python -m SimpleHTTPServer 8080
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# python3 -m http.server 8080
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# 一行代码实现变量值互换
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a, b = 1, 2; a, b = b, a
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# 一行代码解决FizzBuzz问题: 打印数字1到100, 3的倍数打印“Fizz”来替换这个数, 5的倍数打印“Buzz”, 既是3又是5的倍数的打印“FizzBuzz”
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print(' '.join(["fizz"[x % 3 * 4:]+"buzz"[x % 5 * 4:] or str(x) for x in range(1, 101)]))
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# 一行代码输出特定字符"Love"拼成的心形
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print('\n'.join([''.join([('Love'[(x-y) % len('Love')] if ((x*0.05)**2+(y*0.1)**2-1)**3-(x*0.05)**2*(y*0.1)**3 <= 0 else ' ') for x in range(-30, 30)]) for y in range(30, -30, -1)]))
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# 一行代码输出Mandelbrot图像: Mandelbrot图像中的每个位置都对应于公式N=x+y*i中的一个复数
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print('\n'.join([''.join(['*'if abs((lambda a: lambda z, c, n: a(a, z, c, n))(lambda s, z, c, n: z if n == 0 else s(s, z*z+c, c, n-1))(0, 0.02*x+0.05j*y, 40)) < 2 else ' ' for x in range(-80, 20)]) for y in range(-20, 20)]))
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# 一行代码打印九九乘法表
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print('\n'.join([' '.join(['%s*%s=%-2s' % (y, x, x*y) for y in range(1, x+1)]) for x in range(1, 10)]))
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# 一行代码计算出1-100之间的素数(两个版本)
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print(' '.join([str(item) for item in filter(lambda x: not [x % i for i in range(2, x) if x % i == 0], range(2, 101))]))
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print(' '.join([str(item) for item in filter(lambda x: all(map(lambda p: x % p != 0, range(2, x))), range(2, 101))]))
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# 一行代码输出斐波那契数列
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print([x[0] for x in [(a[i][0], a.append([a[i][1], a[i][0]+a[i][1]])) for a in ([[1, 1]], ) for i in range(30)]])
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# 一行代码实现快排算法
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qsort = lambda arr: len(arr) > 1 and qsort(list(filter(lambda x: x <= arr[0], arr[1:]))) + arr[0:1] + qsort(list(filter(lambda x: x > arr[0], arr[1:]))) or arr
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# 一行代码解决八皇后问题
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[__import__('sys').stdout.write('\n'.join('.' * i + 'Q' + '.' * (8-i-1) for i in vec) + "\n========\n") for vec in __import__('itertools').permutations(range(8)) if 8 == len(set(vec[i]+i for i in range(8))) == len(set(vec[i]-i for i in range(8)))]
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# 一行代码实现数组的flatten功能: 将多维数组转化为一维
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flatten = lambda x: [y for l in x for y in flatten(l)] if isinstance(x, list) else [x]
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# 一行代码实现list, 有点类似与上个功能的反功能
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array = lambda x: [x[i:i+3] for i in range(0, len(x), 3)]
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# 一行代码实现求解2的1000次方的各位数之和
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print(sum(map(int, str(2**1000))))
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# 最后推荐一篇文章: [Python One-liner Games](http://arunrocks.com/python-one-liner-games/)
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exit()
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