How can I extract a string from a text field that matches with a “value” in my Dictionary and return the dict’s key into a new column in the Output? [closed] solved How can I extract a string from a text field that matches with a “value” in my Dictionary and return the dict’s key … Read more
You’re resetting list_change to [] on every iteration. Try moving it up to where you define your other global variables. solved I’m Coding A Naughts and Crosses Game [closed]
I would use findall for this.. >>> import re >>> s=”AN : GSHJ488GL67 Customer : sh3893 Acnt No : cgk379gu Name : xyz” >>> re.findall(r’\b(?:AN|Acnt No) : (\w+)’, s) [‘GSHJ488GL67’, ‘cgk379gu’] Explanation: \b # the boundary between a word character and not a word character (?: # group, but do not capture: AN # ‘AN’ … Read more
The problem is that the return true should not happen until the for loop has completed. what we have in the original code is a cuple of tests for trivial cases (x less than 3) and then a loop for testing all the larger numbers. In the loop an attempt is made to divide x … Read more
Something like this? def decode (line): return ‘ ‘.join (token [:-3] if token.endswith (‘WAY’) else (lambda a, b: b [:-3] + a) (*token.split (‘X’) ) for token in line.split () ) Sample: >>> decode (‘elloXHWEY orldXwWEY isXthWEY isWAY eatXgrWEY’) ‘Hello world this is great’ Nota bene: Might fail with words conaining ‘X’. But it should … Read more
I wasn’t able to figure out exactly what you were looking for from your post, but here’s what I think you want: Given a list, get all the numbers at even indices. If any of these numbers are even, put them in a new list and return it: In [10]: L = [3,1,54,5,2,3,4,5,6,5,2,5,3,2,5,2,2,5,2,5,2] In [11]: … Read more
You should calculate consecutive numbers in dedicated variable Try this limit = int(input(“Limit:”)) base = 0 number = 1 while base < limit: base += number number += 1 print(base) 2 solved The Sum of Consecutive Numbers in Python
If all you want to do is return the index of b inside of a, if b is present, and so something else if it is not, try this : – a = [“one”, “two”, “three”, “four”] b = “three” try: print a.index(b) #if inside the function, use return except: #do something2 solved How to … Read more
You could achieve a similar behaviour by using a dictionary and a for-loop iterating through it: #!/usr/bin/env python3 # coding: utf-8 # define a dict with sample data d = {‘fruit’: 13} # print keys and values of sample dict for key, val in d.items(): print(key, val) Output looks like: fruit 13 2 solved How … Read more
If list.sort would return a copy of itself, programmers not knowing that the sorting happens in place would happily write code like this: def foo(my_list): return my_list.sort() Since this works, they might be under the impression that the sorting happens only on a copy of the list, until some day they realise that they’ve been … Read more
#example of the list being used conundrum = [‘black pen’, ‘lift pen’, [‘go to’, -98, 132], [‘draw dot’, 10], ‘lift pen’, [‘go to’, -120, 137], [‘draw dot’, 10], ‘lift pen’, ‘thick lines’, [‘go to’, -55, 80], ‘lower pen’] #Import everything from the turtle library from turtle import * #Define the draw function def draw(test): home() … Read more
Oneliner: list(zip(*l)) # [(1, 3, 5), (2, 4, 6)] 1 solved Printing tuples in a list
Python 2 tries to provide a sort order for almost everything; dictionaries are no exception. Dictionaries are arbitrarily but consistently ordered when compared to one another to ensure that you can sort a heterogenous list that contains them. You should not derive any meaning from their comparisons, really. Python 3 abandoned the notion that all … Read more
Your configurationID is None. This likely means that generate_configurationID() is not returning a value. There is no way in Python for a variable name to “lose” its value. The only way, in the code you posted, for configurationID to be None is for generate_configurationID() to return None which is what will happen if you don’t … Read more
You can use itertools to do this: from itertools import permutations strings = [‘a’,’b’,’c’,’d’,’ab’,’cd’,’abc’] all_combin = [s for i in range(2,len(strings)+1) for s in permutations(strings,i)] # List of all possible combinations of letters from the list num = [] for n in all_combin: if ”.join(n) == ‘abcd’: a=”” for l in n: a += str(strings.index(l)+1) … Read more