import sys
print('Program arguments:', sys.argv)

import report
description = report.get_description()
print("Today's weather:", description)

report.py

def get_description():from random import choice   #可以改成import random      random.choice(possibilities)possibilities = ['rain', 'snow','sleet','fog']return choice(possibilities)

D:\MyPython>python weatherman.py
("Today's weather:", 'snow')

import report as wr
description = wr.get_description()
print("Today's weather:", description)

导入模块的一部分

from report import get_description as do_it
description = do_it()
print("Today's weather:", description)

模糊搜索路径

import sys
for place in sys.path:print(place)

from sources import daily,weekly
print("Daily forecast:", daily.forecast())
print("Weekly forecast:")
for number ,outlook in enumerate(weekly.forecast(),1):    #enumerate()拆分一个列表print(number, outlook)

def forecast():'fake daily forecast'return 'like yesterday'

def forecast():return ['snow', 'more snow', 'sleet', 'freezing rain', 'fog','hail']

('Daily forecast:', 'like yesterday')
Weekly forecast:
(1, 'snow')
(2, 'more snow')
(3, 'sleet')
(4, 'freezing rain')
(5, 'fog')
(6, 'hail')

periodic_table = {'Hydrogen':1,'Helium':2}
carbon = periodic_table.setdefault('Carbon', 12)  #如果键不存在字典中，会被添加进去
print(carbon)
print(periodic_table)
helium = periodic_table.setdefault('Helium', 11)       #试图把一个不同的默认值赋给已经存在的键，不会改变原来的返回原来的值
print(helium)

D:\MyPython>python test2.py
12
{'Helium': 2, 'Hydrogen': 1, 'Carbon': 12}
2

#_*_ coding:utf_8_*_
from collections import defaultdict
periodic_table = defaultdict(int)      #按照int（）调用，返回整数0
periodic_table['Hydrogen'] = 1
print(periodic_table['Lead'])
def no_idea():return 'Huh?'
bestiary = defaultdict(no_idea)
bestiary['A'] = 'Abominable'
bestiary['B'] = 'Basilisk'
print(bestiary['C'])
print(bestiary)

D:\MyPython>python test2.py
0
Huh?
defaultdict(<function no_idea at 0x034FACF0>, {'A': 'Abominable', 'C': 'Huh?', 'B': 'Basilisk'})
用lambda来定义你的默认值函数

# _*_ coding: utf-8_*_
from collections import Counter
breakfast = ['span', 'span', 'eggs', 'span']
breakfast_counter = Counter(breakfast)
print(breakfast_counter)
print(breakfast_counter.most_common())
#降序返回所有元素，或者给定一个数字，返回该数字前的元素
lunch = ['eggs','eggs', 'bacon']
lunch_counter = Counter(lunch)
print(lunch_counter)
#组合计数器的方法是使用 ：+
print(breakfast_counter + lunch_counter)
#什么是早餐有，午餐没有的 ：-
print(breakfast_counter - lunch_counter)
#什么是两者都有的： &
print(breakfast_counter & lunch_counter)
#一共吃了什么东西
print(breakfast_counter | lunch_counter)

# _*_ coding: utf-8_*_
from collections import OrderedDict
quotes = {'Moe': 'A wise', 'Larry': 'Ow!', 'Curly': 'Nyyk'}
for stooge in quotes:print(stooge)
quotes = OrderedDict([('Moe', 'A wise'), ('Larry', 'Ow!'), ('Curly', 'Nyyk')])
for stooge in quotes:print(stooge)

# _*_ coding: utf-8_*_
def palindrome(word):from collections import dequedq = deque(word)while len(dq) > 1:if dq.popleft() != dq.pop():   #popleft()去掉最左边的项并返回，pop()去掉最右边的并返回return Falsereturn True
print(palindrome('a'))

D:\MyPython>python test2.py
True
python没有对字符串进行反转的函数reverse(),但可以利用反向切片

def palindrome(word):return word == word[::-1]
print(palindrome('radar'))

使用itertools迭代代码结构

# _*_ coding: utf-8_*_
import itertools
for item in itertools.chain([1,2],['a', 'b']):print(item)

import itertools
for item in itertools.cycle([1,2]):print(item)

accumulate()计算累积的值，默认的话，它计算的是累加和：

>>> def multiply(a, b):
...     return a * b
...
>>> for item in itertools.accumulate([1,2,3,4],multiply):
...     print(item)
...
1
2
6
24

使用pprint()可以尽量排列输出元素从而增强可读性：

本文是我在学习了丁嘉瑞的《Python语言及其应用》所做的笔记，只用于学习。