When using combinations() you are getting all pairs, even the non-adjacent ones. You can create a function that will return the adjacent pairs. I’ve tried the following code and it worked, maybe it can give you some insight: import os import re from collections import Counter def pairs(text): ans = re.findall(r'[A-Za-z]+’, text) return (tuple(ans[i:i+2]) for … Read more
Its simple, d += ‘a’ + ‘b’ # is same as [] + list(‘ab’) = [‘a’, ‘b’] Similarly d += [‘a’ + ‘b’] # is same as [] + [‘ab’] = [‘ab’] When you type convert a string to list, it creates a list of individual character as elements of list. For Example: a = … Read more
How’s that? from itertools import groupby input_list = [0, 1, 0, 0, 1, 1, 1, 0, 1, 0] results = [0, 0] # First value is zeros, second is ones for key, values in groupby(input_list): results[key] += (2**len(tuple(values))) – 1 assert results == [6,9] solved a pythonic way of doubling consecutive binary bits [closed]
In python? why do the subclass inherit from parent? when you use the init of method to assign the parent?but it does not work [duplicate] solved In python? why do the subclass inherit from parent? when you use the init of method to assign the parent?but it does not work [duplicate]
I can’t say the question is very clear, but if the question is just the title then the answer is simple: 26^6 or in python syntax 26**6 If you want a print statement it would be print(26**6) More generally: The amount of combinations for X characters where any character has N combinations is N^X Do … Read more
a[2] = [3,4] a[2][1] = 4 # second element of the list a[2] a[a[2][1]] = a[4] = 5 Remember that in Python index starts from 0. solved Understanding python list calls
and is how we write conjunctions in Python, so: if msg.topic == “abc” and msg.payload == b’1′: solved Unsupported operand types for &: ‘str’ and ‘bytes’ in python [duplicate]
Assuming that your list is: this_is_a_list = [‘*’, ‘I’, ‘B’, ‘M’, ‘ ‘, ‘i’, ‘s’, ‘ ‘, ‘a’, ‘ ‘, ‘t’, ‘r’, ‘a’, ‘d’, ‘e’, ‘m’, ‘a’, ‘r’, ‘k’, ‘ ‘, ‘o’, ‘f’, ‘ ‘, ‘t’, ‘h’, ‘e’, ‘ ‘, ‘I’, ‘n’, ‘t’, ‘e’, ‘r’, ‘n’, ‘a’, ‘t’, ‘i’, ‘o’, ‘n’, ‘a’, ‘l’, ‘ ‘, … Read more
Or, as another alternative, >>> data=”105.00000031 105.0000041″ >>> print(‘\n’.join(data.split())) 105.00000031 105.0000041 Handles irregular spaces gracefully. 5 solved Modify string representation of data in Python
You need to access the voice channel object. I recommend you use the voice channel’s id. The command could look as follows: @client.command(pass_context = True) async def vcmembers(ctx, voice_channel_id): #First getting the voice channel object voice_channel = discord.utils.get(ctx.message.server.channels, id = voice_channel_id) if not voice_channel: return await client.say(“That is not a valid voice channel.”) members = … Read more
1) Because there is not a clear, unambiguous meaning of str+int. Consider: x = “5” + 7 Should the + str-ify the 7 or int-ify the “5”? One way yields 12, the other yields “57”. 2) Because there are other alternatives that more clearly express the programmer’s intent: print “5”, 7 print “5%d” % 7 … Read more
This appears to be the correct indentation. #!/usr/bin/env python import hashlib import sys def main(): if len(sys.argv) < 2: print “[ + ] Usage: %s <hash>” % sys.argv[0] sys.exit(0) commonStrings = [ “Diaa”, “Diab”, “Mohammad”, “test”, “7amama”, “sos”, “lolo”, “hacked”, “try”, “[email protected]”, “secgeek”, “lnxg33k”, “[email protected]”, “[email protected]”, “[email protected]” ] for i in commonStrings: if hashlib.md5(i).hexdigest() == … Read more
One solution is to use a defaultdict: from collections import defaultdict my_dict = defaultdict(lambda: []) my_dict[‘var1’].append(1) print(my_dict[‘var1’]) # prints ‘[1]’ This would not allow you to simply do print(var1), however, because it would still be undefined in your local (or global) namespace as a tagged value. It would only exist in the defaultdict instance as … Read more
You need for-loop with if/else to run different code for different lists. For list with “PRODUCT NAME PACK” you have to keep list with names in variable – so you may use it in next loops when you get list with numbers For other list you can keep first number as index and rest use … Read more
Since the strings start with index 0, it is incorrect to iterate through the string using while i <= len(summary):, you should use while i < len(summary): instead. Better yet is to use something like: for i in range(len(summary)): solved problem with comparing string with ‘{‘ sign [duplicate]