?(附代碼)分享 | 幾個簡單易懂的Python技巧，極大的提高工作效率

my_string = "ABCDE"reversed_string = my_string[::-1]print(reversed_string)--------------------------------------# Output# EDCBA

my_string = "my name is xiao ming"# 通過title()來實現(xiàn)首字母大寫new_string = my_string.title()print(new_string)-------------------------------------# output# My Name Is Xiao Ming

my_string = "aabbbbbccccddddeeeff"# 通過set()來進(jìn)行去重temp_set = set(my_string)# 通過join()來進(jìn)行連接new_string = ''.join(temp_set)print(new_string)--------------------------------# output# dfbcae

string_1 = "My name is xiao ming"string_2 = "sample, string 1, string 2"
# 默認(rèn)的分隔符是空格，來進(jìn)行拆分print(string_1.split())
# 根據(jù)分隔符"，"來進(jìn)行拆分print(string_2.split(','))------------------------------------# output# ['My', 'name', 'is', 'xiao', 'ming']# ['sample', ' string 1', ' string 2']

list_of_strings = ['My', 'name', 'is', 'Xiao', 'Ming']
# 通過空格和join來連詞成句print(' '.join(list_of_strings))-----------------------------------------# output# My name is Xiao Ming

from collections import Counter
my_list = ['a','a','b','b','b','c','d','d','d','d','d']count = Counter(my_list) print(count) # Counter({'d': 5, 'b': 3, 'a': 2, 'c': 1})
print(count['b']) # 單獨的“b”元素出現(xiàn)的次數(shù)# 3
print(count.most_common(1)) # 出現(xiàn)頻率最多的元素# [('d', 5)]

dict_1 = {'apple': 9, 'banana': 6}dict_2 = {'grape': 4, 'orange': 8}# 方法一combined_dict = {**dict_1, **dict_2}print(combined_dict)# 方法二dict_1.update(dict_2)print(dict_1)# 方法三print(dict(dict_1.items() | dict_2.items()))---------------------------------------# output # {'apple': 9, 'banana': 6, 'grape': 4, 'orange': 8}# {'apple': 9, 'banana': 6, 'grape': 4, 'orange': 8}# {'apple': 9, 'banana': 6, 'grape': 4, 'orange': 8}

import time
start_time = time.time()######################### 具體的程序..........########################end_time = time.time()time_taken_in_micro = (end_time- start_time) * (10 ** 6)print(time_taken_in_micro)

from iteration_utilities import deepflattenl = [[1,2,3],[4,[5],[6,7]],[8,[9,[10]]]]
print(list(deepflatten(l, depth=3)))-----------------------------------------# output# [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

def unique(l):    if len(l)==len(set(l)):        print("不存在重復(fù)值")    else:        print("存在重復(fù)值")
unique([1,2,3,4])# 不存在重復(fù)值
unique([1,1,2,3])# 存在重復(fù)值

array = [['a', 'b'], ['c', 'd'], ['e', 'f']]transposed = zip(*array)print(list(transposed)) ------------------------------------------# output# [('a', 'c', 'e'), ('b', 'd', 'f')]

def difference(a, b):    set_a = set(a)    set_b = set(b)    comparison = set_a.difference(set_b)    return list(comparison)
# 返回第一個列表的不同的元素difference([1,2,6], [1,2,5])# [6]

def to_dictionary(keys, values):    return dict(zip(keys, values))    keys = ["a", "b", "c"]    values = [2, 3, 4]print(to_dictionary(keys, values))-------------------------------------------# output# {'a': 2, 'b': 3, 'c': 4}

d = {'apple': 9, 'grape': 4, 'banana': 6, 'orange': 8}# 方法一sorted(d.items(), key = lambda x: x[1]) # 從小到大排序# [('grape', 4), ('banana', 6), ('orange', 8), ('apple', 9)]sorted(d.items(), key = lambda x: x[1], reverse = True) # 從大到小排序# [('apple', 9), ('orange', 8), ('banana', 6), ('grape', 4)]# 方法二from operator import itemgetterprint(sorted(d.items(), key = itemgetter(1)))# [('grape', 4), ('banana', 6), ('orange', 8), ('apple', 9)]

list1 = [20, 30, 50, 70, 90]
def max_index(list_test):    return max(range(len(list_test)), key = list_test.__getitem__)
def min_index(list_test):    return min(range(len(list_test)), key = list_test.__getitem__)
max_index(list1)# 4min_index(list1)# 0