## How to do recursion precisely in python including lists?

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I am a beginner trying to learn recursion in Python. I want to print all the permutations of a given string. For example:

**Input**: AABC

**Output**: AABC,AACB,ABAC,ABCA,ACAB,BAAC,BACA,BCAA,CAAB,CABA,CBAA

I have written the following code in Python using recursion.

def foo(str,count): result = "" while(any(count)): for i in range(len(count)): if count[i]>0: count[i] -= 1 result = str[i] + foo(str,count) print(result) return result s = "AABC" n = 4 c = {} for i in range(len(s)): if s[i] in c: c[s[i]] += 1 else: c[s[i]] = 1 str = list(c.keys()) count = list(c.values()) print(str,count) foo(str,count) print(count)

I am getting the output as follows:

`['A', 'B', 'C'] [2, 1, 1]`

`C`

`BC`

`ABC`

`AABC`

`[0, 0, 0]`

It implies that the code is handling only the first case at every level. How can I correct this code? Any help would be wonderful. Thanks for your time :)

I'm not sure what you are using `count`

for but here is one way to do it recursively (not very optimal though). :

def foo(s,result): if len(result) == 4: print(result) if len(s) == 0: return for i in range(len(s)): new_s = s.copy() del new_s[i] foo(new_s,result + s[i]) s = list('AABC') foo(s,'')

Output:

AABC AACB ABAC ABCA ACAB ACBA AABC AACB ABAC ABCA ACAB ACBA BAAC BACA BAAC BACA BCAA BCAA CAAB CABA CAAB CABA CBAA CBAA

If you want distinct strings, you can add those to a `set`

**Python Tutorial: Recursive Functions,** A recursion can end up in an infinite loop, if the base case is not met in the calls. It makes sense to define 0! to be 1, because there is exactly one permutation of fibonacci.py, which we can import in the program (fibonacci_runit.py) below: 1-Create a list of integers from two to n: 2, 3, 4, , n 2-Start with a counter i set to� 0 How to do recursion precisely in python including lists? Nov 13 '19. 0 How to pass props through Route without loosing them after refresh? Dec 21 '19.

Your code look a bit messy and it's hard for me to understand how you tried to solve your problem with it. I tried to create as simple solution as possible to help you understand logic behind creating combinations of any kind of list.

def combinations(x): if len(x) == 1: yield x for idx, i in enumerate(x): for comb in combinations(x[:idx]+x[idx+1:]): yield i + comb s = "AABC" print(list(combinations(s))) # -> ['AABC', 'AACB', 'ABAC', 'ABCA', 'ACAB' ...

`x[:idx]+x[idx+1:]`

here is just short of getting rid of `x`

's element at `idx`

's position. Add a comment if you have any questions so I can help you better understand my solution.

**Recursion — How to Think Like a Computer Scientist: Learning with ,** This is a typical example of a problem which is recursive in nature. Most of the Python data types we have seen can be grouped inside lists and tuples in a so that the total size of the subproblem case (including the tear) remains exactly the � At the very least, you should be familiar with Python’s built-in data types, such as lists and tuples. In addition, some familiarity with recursion , classes , data classes , and lambdas will help you better understand the concepts you’ll see in this tutorial.

I figured out something which works but it gives duplicate answers because of multiple occurrences of character 'A', hence I used set() to get the desired answer.

def foo(s,l,result): if len(result)==4: l.append(result) for i in range(len(s)): foo(s[0:i]+s[i+1:],l,result+s[i]) s = "AABC" l = [] r = foo(s,l,"") print(list(set(l)))

**User's Guide, Chapter 5: Lists of Lists, Functions, and Recursion,** In the last Chapter, we discussed Python Lists, how the Stream object is similar can hold all sorts of other things inside them including other lists. First that pi isn't exactly 3.14 – we all know that; I just wanted to make sure� To better understand how recursion works and see it in action using Python, check out Thinking Recursively in Python. Divide-and-conquer algorithms typically follow the same structure: The original input is broken into several parts, each one representing a subproblem that’s similar to the original but simpler.

**Thinking Recursively in Python – Real Python,** Learn how to work with recursion in your Python programs by mastering concepts such as Starting with an empty list, you can generate any list by recursively applying the attach_head Other examples include set, tree, dictionary, etc. 2.3 Lists, Iteration, and Recursion. Racket is a dialect of the language Lisp, whose name originally stood for “LISt Processor.” The built-in list datatype remains a prominent feature of the language.

**Recursion,** def interleave(list1, list2): # assume list1 and list2 are same-length lists if (len(list1 ) == 0): return [] else: return (We need to use "memoization" to speed it up! B) Turn into Python (The "magic" is recursion!) assert(iterativeHanoi(n) == recursiveHanoi(n)) print "iterative and recursive solutions match exactly in all tests! 2.3 Recursion. The idea of calling one function from another immediately suggests the possibility of a function calling itself. The function-call mechanism in Java supports this possibility, which is known as recursion. Your first recursive program.

**ProjPython – Recursion,** It may take you a little while to get comfortable with recursion, however. frame contains a slot for all the local variables (including formal parameters) of that function. If you want a substring of a string (or a sublist of a list), you can “slice” it by using a colon We need to keep track of exactly where each call was made. In this Python dictionaries tutorial you'll cover the basic characteristics and learn how to access and manage dictionary data. Once you have finished this tutorial, you should have a good sense of when a dictionary is the appropriate data type to use, and how to do so.

##### Comments

- Does this answer your question? Python recursion permutations
- What is
`count`

for? - Count is for storing the number of occurrences of each character in string.
- Thanks it works. But I am unfamiliar with enumerate.
- If you're familiar with zip() here's how it could be implemented
`zip(range(len(list)), list)`

. Also enumerate() docs