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Found 10784 Articles for Python
![Pradeep Elance](https://www.tutorialspoint.com/assets/profiles/121076/profile/60_44312-1565945237.jpg)
4K+ Views
Variable assignment is a very basic requirement in any computer programming language. In python there are multiple ways we can declare a variable and assign value to it. Below we see each of them.Direct InitialisationIn this method, we directly declare the variable and assign a value using the = sign. If the variable is declare multiple times, then the last declaration’s value will be used by the program.Examplex = 5 x = 9 print(a)Running the above code gives us the following result:Output9Using if-elseWe can initialize value of a variable using some conditions. The evaluation of the result of the condition ... Read More
![Pradeep Elance](https://www.tutorialspoint.com/assets/profiles/121076/profile/60_44312-1565945237.jpg)
2K+ Views
The universally unique identifier is a 32 bit hexadecimal number that can guarantee a unique value in a given namespace. This helps in tracking down objects created by a program or where ever python needs to handle object or data that needs large value of identifier. The UUID class defines functions that can create these values.Syntaxuuid3(namespace, string) uuid3 usesMD5 hash value to create the identifier. Uuid5(namespace, string) Uuid5 uses SHA-1 hash value to create the identifier. The namespace can be – NAMESPACE_DNS : Used when name string is fully qualified domain name. NAMESPACE_URL : Used when name string is ... Read More
![Pavitra](https://www.tutorialspoint.com/articles/images/user_icon.png)
510 Views
In this article, we will learn about the solution to the problem statement given below.Problem statement − We are given a string we need to display all the possible permutations of the string.Now let’s observe the solution in the implementation below −Example Live Demo# conversion def toString(List): return ''.join(List) # permutations def permute(a, l, r): if l == r: print (toString(a)) else: for i in range(l, r + 1): a[l], a[i] = a[i], a[l] permute(a, l + 1, r) a[l], ... Read More
![Pavitra](https://www.tutorialspoint.com/articles/images/user_icon.png)
599 Views
In this article, we will learn about the solution to the problem statement given below.Problem statement − We are given two strings, we need to get the uncommon words from the given strings.Now let’s observe the solution in the implementation below −Example Live Demo# uncommon words def find(A, B): # count count = {} # insert in A for word in A.split(): count[word] = count.get(word, 0) + 1 # insert in B for word in B.split(): count[word] = count.get(word, 0) + 1 # return ans return [word ... Read More
![Pavitra](https://www.tutorialspoint.com/articles/images/user_icon.png)
754 Views
In this article, we will learn about the solution to the problem statement given below.Problem statement − We are given a function, we need to display the number of local variables in the function.Now let’s observe the solution in the implementation below −Example Live Demo# checking locals def scope(): a = 25.5 b = 5 str_ = 'Tutorialspoint' # main print("Number of local varibales available:", scope.__code__.co_nlocals)OutputNumber of local varibales available: 3Example Live Demo# checking locals def empty(): pass def scope(): a, b, c = 9, 23.4, True str = 'Tutiorialspoint' # main print("Number of local varibales ... Read More
![Pavitra](https://www.tutorialspoint.com/articles/images/user_icon.png)
668 Views
In this article, we will learn about the solution to the problem statement given below.Problem statement − We are given an integer n, we need to count the number of trailing zeros in the factorial.Now let’s observe the solution in the implementation below −Example Live Demo# trailing zero def find(n): # Initialize count count = 0 # update Count i = 5 while (n / i>= 1): count += int(n / i) i *= 5 return int(count) # Driver program n = 79 print("Count of trailing 0s "+"in", n, ... Read More
![Pavitra](https://www.tutorialspoint.com/articles/images/user_icon.png)
330 Views
In this article, we will learn about the solution to the problem statement given below.Problem statement − We are given an integer n, we need to count the number of 1’s in the binary representation of the numberNow let’s observe the solution in the implementation below −#naive approachExample Live Demo# count the bits def count(n): count = 0 while (n): count += n & 1 n >>= 1 return count # main n = 15 print("The number of bits :", count(n))OutputThe number of bits : 4#recursive approachExample Live Demo# recursive way def count( ... Read More
![Pavitra](https://www.tutorialspoint.com/articles/images/user_icon.png)
215 Views
In this article, we will learn about the solution to the problem statement given below.Problem statement − We are given a positive integer N, we need to count all possible distinct binary strings available with length N such that no consecutive 1’s exist in the string.Now let’s observe the solution in the implementation below −Example Live Demo# count the number of strings def countStrings(n): a=[0 for i in range(n)] b=[0 for i in range(n)] a[0] = b[0] = 1 for i in range(1, n): a[i] = a[i-1] + b[i-1] b[i] = ... Read More
![Pradeep Elance](https://www.tutorialspoint.com/assets/profiles/121076/profile/60_44312-1565945237.jpg)
9K+ Views
As a part of text analytics, we frequently need to count words and assign weightage to them for processing in various algorithms, so in this article we will see how we can find the frequency of each word in a given sentence. We can do it with three approaches as shown below.Using CounterWe can use the Counter() from collections module to get the frequency of the words. Here we first apply the split() to generate the words from the line and then apply the most_common ().Example Live Demofrom collections import Counter line_text = "Learn and practice and learn to practice" freq ... Read More
![Pavitra](https://www.tutorialspoint.com/articles/images/user_icon.png)
210 Views
In this article, we will learn about the solution to the problem statement given below.Problem statement − We are given a list, we need to count the inversion required and display it.Inversion count is obtained by counting how many steps are needed for the array to be sorted.Now let’s observe the solution in the implementation below −Example Live Demo# count def InvCount(arr, n): inv_count = 0 for i in range(n): for j in range(i + 1, n): if (arr[i] > arr[j]): inv_count += 1 return ... Read More