re模組補充

findall與分組

findall預設是分組優先展示。正則表示式中如果有括號分組，預設只展示括號內正則表示式匹配到的內容
也可以用(?:)取消分組優先展示的機制

import re
res = re.findall('n(o) ', "The wise man builds no hopes for the future, entertains no regrets for the past")
print(res)  # ['o', 'o']
res = re.findall('n(?:o) ', "The wise man builds no hopes for the future, entertains no regrets for the past")
print(res)  # ['no ', 'no ']
res = re.findall('(n)(o)( )', "The wise man builds no hopes for the future, entertains no regrets for the past")
print(res)   # [('n', 'o', ' '), ('n', 'o', ' ')]
res = re.findall('(?P<name1>n)(o)( )', "The wise man builds no hopes for the future, entertains no regrets for the past")
print(res)   # [('n', 'o', ' '), ('n', 'o', ' ')] 起別名沒有意義，不支援group呼叫
# print(res.group('name1'))  # 'list' object has no attribute 'group'
 

search與分組

# 針對search有多少個分組,group方法括號內參數最大就是多少
res = re.search('n(o) ', "The wise man builds no hopes for the future, entertains no regrets for the past")
print(res.group())  # no
print(res.group(0))  # no
print(res.group(1))  # o
res = re.search('n(o)( )', "The wise man builds no hopes for the future, entertains no regrets for the past")
print(res.group())  # no
print(res.group(0))  # no
print(res.group(1))  # o
print(res.group(2))  # 空格
# 分組之後還可以給組起別名
res = re.search('(?P<name1>n)(?P<name2>o)( )', "The wise man builds no hopes for the future, entertains no regrets for the past")
print(res.group('name1'))  # n
print(res.group('name2'))  # o
 

match與分組

# 針對match有多少個分組,group方法括號內參數最大就是多少
res = re.match('(T)he', "The wise man builds no hopes for the future, entertains no regrets for the past")
print(res.group())  # The
print(res.group(0))  # The
print(res.group(1))  # T
res = re.match('(T)(h)e', "The wise man builds no hopes for the future, entertains no regrets for the past")
print(res.group())  # The
print(res.group(0))  # The
print(res.group(1))  # T
print(res.group(2))  # h
# 分組之後還可以給組起別名
res = re.match('(?P<name1>T)(?P<name2>h)(e)', "The wise man builds no hopes for the future, entertains no regrets for the past")
print(res.group('name1'))  # T
print(res.group('name2'))  # h
 

常用內建模組

collections模組

具名元組

from collections import namedtuple

# 第一步、先產生一個元組物件模板
point = namedtuple('商品', ['product_name', 'product_price'])
# 第二步、建立諸多元組資料
product_info = point('pen', '$50')
product_info1 = point('book', '$20')
product_info2 = point('desk', '$100')
print(product_info, product_info1, product_info2)  # 商品(product_name='pen', product_price='$50') 商品(product_name='book', product_price='$20') 商品(product_name='desk', product_price='$100')
print(product_info.product_name)  # pen
print(product_info.product_price)  # $50
print(product_info2.product_name)  # desk
print(product_info2.product_price)  # $100

雙端佇列

# 普通佇列
import queue
name_info = queue.Queue(5)  # 最大隻能放三個元素
# 存放元素
name_info.put('jack')
name_info.put('tom')
name_info.put('annie')
name_info.put('sam')
name_info.put('mary')
# name_info.put('jerry')  # 佇列以滿,原地等待
# 獲取元素
print(name_info.get())  # jack
print(name_info.get())  # tom
print(name_info.get())  # annie
print(name_info.get())  # sam
print(name_info.get())  # mary
# print(name_info.get())  # 佇列以空，原地等待

# 雙端佇列
from collections import deque

name_info = deque(['jack', 'tom', 'annie'])
print(name_info)  # deque(['jack', 'tom', 'annie'])
name_info.append('sam')  # 右邊新增元素
print(name_info)  # deque(['jack', 'tom', 'annie', 'sam'])
name_info.appendleft('mary')  # 左邊新增元素
print(name_info)  # deque(['mary', 'jack', 'tom', 'annie', 'sam'])
name_info.pop()  # 右邊彈出元素
name_info.popleft()  # 左邊彈出元素
print(name_info)  # deque(['jack', 'tom', 'annie'])

字典相關

# 正常的字典內部是無序的
d1 = dict([('name', 'sam'), ('age', 20), ('hobby', 'play games'), ('height', '175cm')])
for i in d1:
    print(i)  # name age hobby height
print(d1)  # {'name': 'sam', 'age': 20, 'hobby': 'play games', 'height': '175cm'}
print(d1.keys())  # dict_keys(['name', 'age', 'hobby', 'height'])
# 有序字典
from collections import OrderedDict

d2 = OrderedDict([('a', 97), ('b', 98), ('c', 99)])
print(d2)  # OrderedDict([('a', 97), ('b', 98), ('c', 99)])
print(chr(102))  # f
d2['d'] = 100
d2['e'] = 101
d2['f'] = 102
print(d2)  # OrderedDict([('a', 97), ('b', 98), ('c', 99), ('d', 100), ('e', 101), ('f', 102)])
print(d2.keys())  # odict_keys(['a', 'b', 'c', 'd', 'e', 'f'])

將列表中[11, 22, 33, 44, 55, 66, 77, 88, 99, 111, 222, 333, 444, 555, 666]大於111的值儲存到字典的第一個key中，將不大於111的值儲存至第二個key的值中。

l1 = [11, 22, 33, 44, 55, 66, 77, 88, 99, 111, 222, 333, 444, 555, 666]
dict1 = {'k1': [], 'k2': []}
for i in l1:
    if i > 111:
        dict1['k1'].append(i)
    else:
        dict1['k2'].append(i)
print(dict1)  # {'k1': [222, 333, 444, 555, 666], 'k2': [11, 22, 33, 44, 55, 66, 77, 88, 99, 111]}
from collections import defaultdict

l2 = [11, 22, 33, 44, 55, 66, 77, 88, 99, 111, 222, 333, 444, 555, 666]
dict2 = defaultdict(list)
for value in l2:
    if value > 111:
        dict2['k1'].append(value)
    else:
        dict2['k2'].append(value)
print(dict2)  # defaultdict(<class 'list'>, {'k2': [11, 22, 33, 44, 55, 66, 77, 88, 99, 111], 'k1': [222, 333, 444, 555, 666]})

計數器

string = 'aasdhjbhsBHjababbcbdabbbahabbcajjcbahdhbchccajjcaajcahchahccaasbb'
# 統計字串中所有字元出現的次數
dict1 = {}
for i in string:
    if i not in dict1:
        dict1[i] = 1
    else:
        dict1[i] += 1
print(dict1)  # {'a': 16, 's': 3, 'd': 3, 'h': 9, 'j': 7, 'b': 14, 'B': 1, 'H': 1, 'c': 11}
from collections import Counter

res = Counter(string)
print(res)  # Counter({'a': 16, 'b': 14, 'c': 11, 'h': 9, 'j': 7, 's': 3, 'd': 3, 'B': 1, 'H': 1})
print(res.get('a'))  # 16
print(res.get('b'))  # 14
print(res.get('j'))  # 7