Define your function example as this:- def exemple(func_name): def dostuff1(): print(‘Stuff 1’) def dostuff2(): print(‘Stuff 2’) def dostuff3(): print(‘Stuff 3’) func_dic = { “dostuff1” : dostuff1, “dostuff2” : dostuff2, “dostuff3” : dostuff3 } return func_dic[func_name] Then call your function in the other function like this: def training(): exemple(“dostuff2”)() I hope it helps! 3 solved Python … Read more
I wonder how you are able to run this program, seems to be some harry potter stuff. For me its typo in this line “ele” == input(“”) and gives Syntax error . So just fix this with ele = input() and replace “ele” with ele in all if/elif checks i.e if ele == “H”:it will … Read more
Keeping to your original idea, you were fairly close. Note, Python already keeps a handy list of lower case letters: import string alphabet = string.ascii_lowercase message = input(‘Please insert the message you want to encrypt: ‘) key = int(input(‘What key value do you want in your encryption? ‘)) output = [] for m in message: … Read more
Writing a function to print letters of a string according to their frequency compalsarily using dictionary and tuple [closed] solved Writing a function to print letters of a string according to their frequency compalsarily using dictionary and tuple [closed]
to take a year from current date: import datetime year = datetime.datetime.today().year 1 solved How to exclude year from date? [closed]
a==14 is not a statement, it’s an expression. It produces a boolean result, that is discarded as soon as it is produced. It is a line that has absolutely no effect on the program. On the other hand, there is no === operator in python, that’s why your program fails in that case. Contrary to … Read more
sentence = “Get a sentence from the user and display it back with one word per line.” for character in range(len(sentence)): if sentence[character] == ‘ ‘ or sentence[character] == ‘.’ or sentence[character] == ‘,’: print(“”) else: print(sentence[character], end=”) OUTPUT: Get a sentence from the user and display it back with one word per line solved … Read more
Count how many of each alphabetic character. Use a dictionary: It is best to think of a dictionary as an unordered set of key: value pairs, with the requirement that the keys are unique (within one dictionary). A pair of braces creates an empty dictionary: {}. Placing a comma-separated list of key:value pairs within the … Read more
If it was for i in range(1,10): if space_check(board, i): return False else: return True then after the first iteration in the for loop the function would return. This would not lead the the expected behaviour. Currently, you check every space, and not just the first one 9 solved Are booleans overwritten in python?
The problem is, that your value i is an integer and the print-function takes a string as parameter. Therefore you need to cast your variable using the str() function. Your code should look like this: for i in range(999999999999999): print (“sending messages no.” + str(i)) Now, I encourage people to learn Python and so I … Read more
it allows to define a criterion which replaces the < comparison between elements. For instance: >>>l = [“hhfhfhh”,”xx”,”123455676883″] >>>max(l, key=len) ‘123455676883’ returns the longest string in the list which is “123455676883” Without it, it would return “xx” because it’s the highest ranking string according to string comparison. >>>l = [“hhfhfhh”,”xx”,”123455676883″] >>>max(l) ‘xx’ 2 solved What … Read more
To check each digit you need to convert these number in string-object and then count each digit from other number which result 0 or some finite number and then using all we can check that all digit is present their or not. Check out this code: def same_digits(a: int, b: int)->bool: cond1 = all([str(b).count(i) for … Read more
You can use itertools.groupby to group everything into lists of adjacent identical values, then unpack the lists of length one out of the nested lists: from itertools import groupby def group_adjacent(iterable): lists = (list(g) for k, g in groupby(iterable)) return [subl[0] if len(subl) == 1 else subl for subl in lists] group_adjacent([1,2,4,5,5,7,6,6,6]) # [1, 2, … Read more
You can cast it on input like this: hrs = int(input(“Enter Hours:”)) However, there is no need to do this as the compiler can tell which data type has been input. Example: >>> hrs = input(“Enter Hours:”) Enter Hours:30 >>> type(hrs) <type ‘int’> You can see that python already knows what type has been entered. … Read more
your first question: [i for i in range(n) for j in range(5)] your second question: [j for i in range(k) for j in range(1,6)] where k is the number of repeats 5 solved How to create the following lists with list comprehensions? [closed]