my_string = "aavvccccddddeee" 
# 将字符串转换为集合 
temp_set = set(my_string) 
# 使用join将拼接设置为字符串 
new_string = ''.join(temp_set) 
print(new_string) 

my_string = “My Name Is Chaitanya Baweja” 
＃使用string类的title()函数 
new_string = my_string.title() 
print(new_string) 
＃output 
My Name Is Chaitanya Baweja 

#使用切片反转字符串 
my_string = "ABCDE" 
reversed_string = my_string [::-1] 
print(reversed_string) 
#output 

n = 3 #重复次数 
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] 

n = 4 
my_list = [0] * n #n表示所需列表的长度 
#[0,0,0,0] 

list_of_strings = ['My', 'name', 'is', 'haha', 'lulu'] 
# 使用","连接字符串 
print(','.join(list_of_strings)) 
# Output 
# My,name,is,haha,lulu 

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

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

# 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] 

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

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 

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') 

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") 

# 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)] 

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

 随机导入 
my_list = [ 'a'，'b'，'c'，'d'，'e' ] 
num_samples = 2 
样本= 随机 .sample（my_list，num_samples） 
 打印（样本） 
#[ 'a'，'e' ] 这将具有任意2个 随机值

import time 
start_time = time.time() 
# Code to check follows 
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_seconds: {0} ms").format(time_taken_in_micro) 

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] 

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}