Verwenden Sie Python-Slicing, um den String umzukehren:
# Reversing a string using slicing my_string = "ABCDE" reversed_string = my_string[::-1] print(reversed_string) # Output # EDCBA
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
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
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. Listengenerierung
# 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]
a = 1 b = 2 a, b = b, a print(a) # 2 print(b) # 1
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']
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
my_string = "abcba" if my_string == my_string[::-1]: print("palindrome") else: print("not palindrome") # Output # palindrome
# 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)]
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
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
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
import sys num = 21 print(sys.getsizeof(num)) # In Python 2, 24 # In Python 3, 28
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}
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
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]
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
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]
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