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Use Python slicing to reverse the string:
# Reversing a string using slicing my_string = "ABCDE" reversed_string = my_string[::-1] print(reversed_string) # Output # EDCBA
Use the title function Method:
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
Use the concept of sets to find unique elements in strings:
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
You can print a string or list multiple times using the multiplication sign (*):
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]
# 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
Use the .split() function:
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)]
The meaning of Anagrams If each English word (excluding uppercase and lowercase) appears the same number of times in the two words, use the Counter class to determine whether the two strings are 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
except to get exception handling:
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}
Use the time class to calculate Time spent running a piece of code:
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
Convert integers to List of numbers:
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]
Check whether each element in the list is unique:
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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