首先你要了解一下Python之禅,一行代码输出“The Zen of Python”:
python -c "import this"
"""
The Zen of Python, by Tim Peters
Beautiful is better than ugly.
Explicit is better than implicit.
Simple is better than complex.
Complex is better than complicated.
Flat is better than nested.
Sparse is better than dense.
Readability counts.
Special cases aren't special enough to break the rules.
Although practicality beats purity.
Errors should never pass silently.
Unless explicitly silenced.
In the face of ambiguity, refuse the temptation to guess.
There should be one-- and preferably only one --obvious way to do it.
Although that way may not be obvious at first unless you're Dutch.
Now is better than never.
Although never is often better than *right* now.
If the implementation is hard to explain, it's a bad idea.
If the implementation is easy to explain, it may be a good idea.
Namespaces are one honking great idea -- let's do more of those!
"""
从“The Zen of Python”也能看出,Python倡导Beautiful、Explicit、Simple等原则,当然我们接下来要介绍的一行Python能实现哪些好玩的功能,可能和Explicit原则相违背。
(1)一行代码启动一个Web服务
python -m SimpleHTTPServer 8080 # python2
python3 -m http.server 8080 # python3
(2)一行代码实现变量值互换
a, b = 1, 2; a, b = b, a
(3)一行代码解决FizzBuzz问题:
FizzBuzz问题:打印数字1到100, 3的倍数打印“Fizz”, 5的倍数打印“Buzz”, 既是3又是5的倍数的打印“FizzBuzz”
for x in range(1, 101): print("fizz"[x % 3 * 4:]+"buzz"[x % 5 * 4:] or x)
(4)一行代码输出特定字符”Love”拼成的心形
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)]))
Mandelbrot图像:图像中的每个位置都对应于公式N=x+y*i中的一个复数
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)]))
print('\n'.join([' '.join(['%s*%s=%-2s' % (y, x, x*y) for y in range(1, x+1)]) for x in range(1, 10)]))
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))]))
print(' '.join([str(item) for item in filter(lambda x: all(map(lambda p: x % p != 0, range(2, x))), range(2, 101))]))
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)]])
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
[__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)))]
flatten = lambda x: [y for l in x for y in flatten(l)] if isinstance(x, list) else [x]
array = lambda x: [x[i:i+3] for i in range(0, len(x), 3)]
(13)一行代码实现求解2的1000次方的各位数之和
print(sum(map(int, str(2**1000))))