Python初学者的18个技巧

交换变量

x = 6

y = 5

x, y = y, x

print x

>>> 5

print y

>>> 6

if 语句在行内

print "Hello" if True else "World"

>>> Hello

连接

下面的最后一种方式在绑定两个不同类型的对象时显得很酷。

nfc = ["Packers", "49ers"]

afc = ["Ravens", "Patriots"]

print nfc + afc

>>> ['Packers', '49ers', 'Ravens', 'Patriots']

print str(1) + " world"

>>> 1 world

print `1` + " world"

>>> 1 world

print 1, "world"

>>> 1 world

print nfc, 1

>>> ['Packers', '49ers'] 1

计算技巧

#向下取整

print 5.0//2

>>> 2

# 2的5次方

print 2**5

>> 32

注意浮点数的除法

print .3/.1

>>> 2.9999999999999996

print .3//.1

>>> 2.0

数值比较

x = 2

if 3 > x > 1:

print x

>>> 2

if 1 < x > 0:

print x

>>> 2

两个列表同时迭代

nfc = ["Packers", "49ers"]

afc = ["Ravens", "Patriots"]

for teama, teamb in zip(nfc, afc):

print teama + " vs. " + teamb

>>> Packers vs. Ravens

>>> 49ers vs. Patriots

带索引的列表迭代

teams = ["Packers", "49ers", "Ravens", "Patriots"]

for index, team in enumerate(teams):

print index, team

>>> 0 Packers

>>> 1 49ers

>>> 2 Ravens

>>> 3 Patriots

列表推导

已知一个列表，刷选出偶数列表方法：

numbers = [1,2,3,4,5,6]

even = []

for number in numbers:

if number%2 == 0:

even.append(number)

用下面的代替

numbers = [1,2,3,4,5,6]

even = [number for number in numbers if number%2 == 0]

字典推导

teams = ["Packers", "49ers", "Ravens", "Patriots"]

print {key: value for value, key in enumerate(teams)}

>>> {'49ers': 1, 'Ravens': 2, 'Patriots': 3, 'Packers': 0}

初始化列表的值

items = [0]*3

print items

>>> [0,0,0]

将列表转换成字符串

teams = ["Packers", "49ers", "Ravens", "Patriots"]

print ", ".join(teams)

>>> 'Packers, 49ers, Ravens, Patriots'

从字典中获取元素

不要用下列的方式

data = {'user': 1, 'name': 'Max', 'three': 4}

try:

is_admin = data['admin']

except KeyError:

is_admin = False

替换为

data = {'user': 1, 'name': 'Max', 'three': 4}

is_admin = data.get('admin', False)

获取子列表

x = [1,2,3,4,5,6]

#前3个

print x[:3]

>>> [1,2,3]

#中间4个

print x[1:5]

>>> [2,3,4,5]

#最后3个

print x[-3:]

>>> [4,5,6]

#奇数项

print x[::2]

>>> [1,3,5]

#偶数项

print x[1::2]

>>> [2,4,6]

60个字符解决FizzBuzz

前段时间Jeff Atwood 推广了一个简单的编程练习叫FizzBuzz，问题引用如下：

写一个程序，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”，对于既是3的倍数又是5的倍数的数字打印“FizzBuzz”。

这里有一个简短的方法解决这个问题：

for x in range(101):print"fizz"[x%3*4::]+"buzz"[x%5*4::]or x

集合

用到Counter库

from collections import Counter

print Counter("hello")

>>> Counter({'l': 2, 'h': 1, 'e': 1, 'o': 1})

迭代工具

和collections库一样，还有一个库叫itertools

from itertools import combinations

teams = ["Packers", "49ers", "Ravens", "Patriots"]

for game in combinations(teams, 2):

print game

>>> ('Packers', '49ers')

>>> ('Packers', 'Ravens')

>>> ('Packers', 'Patriots')

>>> ('49ers', 'Ravens')

>>> ('49ers', 'Patriots')

>>> ('Ravens', 'Patriots')

False == True

在python中，True和False是全局变量，因此：

False = True

if False:

print "Hello"

else:

print "World"

>>> Hello