Well, you got what you told to print! board is a list of list of strs, so board[i] must be a list of strs, and when you write print(board[i]), you get a list! You may need to write this: print(”.join(board[i])) solved What i did wrong in my python function?
It’s ugly, but yeah, it can be done: my_list=[[‘Hello’, 1 , 3],[‘Hi’, 4 , 6]] print(list(map(lambda lst: list(map(lambda x: float(x) if type(x) == int else x, lst)), my_list))) # Output: [[‘Hello’, 1.0, 3.0], [‘Hi’, 4.0, 6.0]] EDIT I personally prefer list comprehension in most cases as opposed to map/lambda, especially here where all the conversions … Read more
You can capture a single frame by using the VideoCapture method of OpenCV. import cv2 pic = cv2.VideoCapture(0) # video capture source camera (Here webcam of laptop) ret,frame = pic.read() # return a single frame in variable `frame` while(True): cv2.imshow(‘img1’,frame) #display the captured image if cv2.waitKey(1) & 0xFF == ord(‘y’): #save on pressing ‘y’ cv2.imwrite(‘images/c1.png’,frame) … Read more
Loop over the dictionary’s items and use replace() over txt for key, val in dict.items(): txt = txt.replace(key, val) Note: it is not a good idea to use dict as a variable name since it is a builtin type in python 1 solved Python: Replace different words in a string using matching array values
username = [‘admin’,’bill’,’Kevin’,’mike’,’nick’] while True : name =input(“Please enter a username: “) if name==’admin’ : print(“Hello “+ name + ” ,would you like to see a status report?”) break elif name in username : print(“Hello ” + name.title() + ” thank you for logging in!”) break else: print(“Who are you ” + name.title() + ” … Read more
You don’t need to pass category_name, you only need the category id. The category name should be contained in each of the item fetched from the database. You’re getting an error because cat_id is not defined when the function def getCategoryItems(category_name, cat_id) is called. I would suggest, however, if you want to really get all … Read more
In order to give a complete answer as to why and what is happening here, first we need to talk about the concept of “scope”. Each function (and for loop, while loop etc) “lives in its own world” in terms of memory. What does that mean? Assume I have a function named “foo1” and another … Read more
list1 is a list that contains 1 element that is an numpy.array that contains multiple floats64. list2 is list that contains 1 element that is a list that contains multiple strings (that happen to look a lot like floats). You can convert them like so: import numpy as np # list of list of strings … Read more
For DataFrames, the value to be passed for axis is columns. frame.sort_index(axis=”columns”) 1 solved No axis named 1 for object type
If I was developing this, I would: Keep all the words in a single set Skip the first question (you can determine word length by the length of the 2nd question) Set up a while loop for each question you ask the user so that it repeats the same question on invalid input. To check … Read more
I don`t really understand the question, but if you want the highest number of consecutive elements in list, maybe something like this from itertools import groupby list = [1,1,1,0,0,1,1,1,1,1] count_cons_val = [sum(1 for _ in group) for _, group in groupby(list)] print(max(count_cons_val)) Output: 5 solved Highest number of consecutively repeating values in a list
To get your message, use: a[0] To get your IP, use: a[1][0] To get 34343 (whatever that is), use: a[1][1] solved How to extract IP address from given tuple with mix of string and integer number
import re pat = [‘ashu’, ‘gupta’] text = “my name is ashutosh gupta” for pattern in pat: print(‘i am seraching for: ‘+pattern) if re.search(pattern, text): print(“Match”) else: print(“NOT mactch”) solved Regular rexpression in Python
The second time, hi is in quotes. So it is a string literal. It is not a variable. Specifically, it is not the value of the iterator. 0 solved Can someone please tell me why the two python codes below produce entirely different outcomes?
dictionary = {‘key’: [‘value’]} print(dictionary.get(‘key’)[0]) solved Get raw key value