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你需要掌握的20個(gè)Python常用技巧

更新時(shí)間：2022年02月28日 14:36:00 作者：愛(ài)摸魚(yú)的菜鳥(niǎo)碼農(nóng)

Python的可讀性和簡(jiǎn)單性是其廣受歡迎的兩大原因，本文介紹20個(gè)常用的Python技巧來(lái)提高代碼的可讀性，并能幫助你節(jié)省大量時(shí)間，需要的可以參考一下

Python的可讀性和簡(jiǎn)單性是其廣受歡迎的兩大原因，本文介紹20個(gè)常用的Python技巧來(lái)提高代碼的可讀性，并能幫助你節(jié)省大量時(shí)間，下面的技巧將在你的日常編碼練習(xí)中非常實(shí)用。

1.字符串反轉(zhuǎn)

使用Python切片反轉(zhuǎn)字符串：

# Reversing a string using slicing
 
my_string = "ABCDE"
reversed_string = my_string[::-1]
 
print(reversed_string)
 
# Output
# EDCBA

2.每個(gè)單詞的第一個(gè)字母大寫(xiě)

使用title函數(shù)方法：

my_string = "my name is chaitanya baweja"
 
# using the title() function of string class
new_string = my_string.title()
 
print(new_string)
 
# Output
# My Name Is Chaitanya Baweja

3. 字符串查找唯一元素

使用集合的概念查找字符串的唯一元素:

my_string = "aavvccccddddeee"
 
# converting the string to a set
temp_set = set(my_string)
 
# stitching set into a string using join
new_string = ''.join(temp_set)
 
print(new_string)
 
# output
# cdvae

4.重復(fù)打印字符串和列表n次

你可以使用乘法符號(hào)（*）打印字符串或列表多次：

n = 3 # number of repetitions
 
my_string = "abcd"
my_list = [1,2,3]
 
print(my_string*n)
# abcdabcdabcd
 
print(my_list*n)
# [1,2,3,1,2,3,1,2,3]

5.列表生成

# Multiplying each element in a list by 2
 
original_list = [1,2,3,4]
 
new_list = [2*x for x in original_list]
 
print(new_list)
# [2,4,6,8]

6.變量交換

a = 1
b = 2
 
a, b = b, a
 
print(a) # 2
print(b) # 1

7.字符串拆分為子字符串列表

使用.split()函數(shù)：

string_1 = "My name is Chaitanya Baweja"
string_2 = "sample/ string 2"
 
# default separator ' '
print(string_1.split())
# ['My', 'name', 'is', 'Chaitanya', 'Baweja']
 
# defining separator as '/'
print(string_2.split('/'))
# ['sample', ' string 2']

8.多個(gè)字符串組合為一個(gè)字符串

list_of_strings = ['My', 'name', 'is', 'Chaitanya', 'Baweja']
 
# Using join with the comma separator
print(','.join(list_of_strings))
 
# Output
# My,name,is,Chaitanya,Baweja

9.檢測(cè)字符串是否為回文

my_string = "abcba"
 
if my_string == my_string[::-1]:
    print("palindrome")
else:
    print("not palindrome")
 
# Output
# palindrome

10. 統(tǒng)計(jì)列表中元素的次數(shù)

# finding frequency of each element in a list
from collections import Counter
 
my_list = ['a','a','b','b','b','c','d','d','d','d','d']
count = Counter(my_list) # defining a counter object
 
print(count) # Of all elements
# Counter({'d': 5, 'b': 3, 'a': 2, 'c': 1})
 
print(count['b']) # of individual element
# 3
 
print(count.most_common(1)) # most frequent element
# [('d', 5)]

11.判斷兩個(gè)字符串是否為Anagrams

Anagrams的含義為兩個(gè)單詞中，每個(gè)英文單詞（不含大小寫(xiě)）出現(xiàn)的次數(shù)相同，使用Counter類判斷兩個(gè)字符串是否為Anagrams。

from collections import Counter
 
str_1, str_2, str_3 = "acbde", "abced", "abcda"
cnt_1, cnt_2, cnt_3  = Counter(str_1), Counter(str_2), Counter(str_3)
 
if cnt_1 == cnt_2:
    print('1 and 2 anagram')
if cnt_1 == cnt_3:
    print('1 and 3 anagram')
 
# output
# 1 and 2 anagram

12. 使用try-except-else-block模塊

except獲取異常處理：

a, b = 1,0
 
try:
    print(a/b)
    # exception raised when b is 0
except ZeroDivisionError:
    print("division by zero")
else:
    print("no exceptions raised")
finally:
    print("Run this always")
 
# output
# division by zero
# Run this always

13. 使用枚舉函數(shù)得到key/value對(duì)

my_list = ['a', 'b', 'c', 'd', 'e']
 
for index, value in enumerate(my_list):
    print('{0}: {1}'.format(index, value))
 
# 0: a
# 1: b
# 2: c
# 3: d
# 4: e

14.檢查對(duì)象的內(nèi)存使用情況

import sys
 
num = 21
 
print(sys.getsizeof(num))
 
# In Python 2, 24
# In Python 3, 28

15.合并字典

dict_1 = {'apple': 9, 'banana': 6}
dict_2 = {'banana': 4, 'orange': 8}
 
combined_dict = {**dict_1, **dict_2}
 
print(combined_dict)
# Output
# {'apple': 9, 'banana': 4, 'orange': 8}

16.計(jì)算執(zhí)行一段代碼所花費(fèi)的時(shí)間

使用time類計(jì)算運(yùn)行一段代碼所花費(fèi)的時(shí)間：

import time
 
start_time = time.time()
# Code to check follows
for i in range(10**5):
    a, b = 1,2
    c = a+ b
# Code to check ends
end_time = time.time()
time_taken_in_micro = (end_time- start_time)*(10**6)
 
print(time_taken_in_micro)
 
# output
# 18770.217895507812

17. 列表展開(kāi)

from iteration_utilities import deepflatten
 
# if you only have one depth nested_list, use this
def flatten(l):
  return [item for sublist in l for item in sublist]
 
l = [[1,2,3],[3]]
print(flatten(l))
# [1, 2, 3, 3]
 
# if you don't know how deep the list is nested
l = [[1,2,3],[4,[5],[6,7]],[8,[9,[10]]]]
 
print(list(deepflatten(l, depth=3)))
# [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

18. 列表采樣

import random
 
my_list = ['a', 'b', 'c', 'd', 'e']
num_samples = 2
 
samples = random.sample(my_list,num_samples)
print(samples)
# [ 'a', 'e'] this will have any 2 random values

19.數(shù)字化

將整數(shù)轉(zhuǎn)化成數(shù)字列表

num = 123456
 
# using map
list_of_digits = list(map(int, str(num)))
 
print(list_of_digits)
# [1, 2, 3, 4, 5, 6]
 
# using list comprehension
list_of_digits = [int(x) for x in str(num)]
 
print(list_of_digits)
# [1, 2, 3, 4, 5, 6]

20.檢查列表元素的唯一性

檢查列表中每個(gè)元素是否為唯一的：

def unique(l):
    if len(l)==len(set(l)):
        print("All elements are unique")
    else:
        print("List has duplicates")
 
unique([1,2,3,4])
# All elements are unique
 
unique([1,1,2,3])
# List has duplicates

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