It’s a dictonary, not a list of strings. I think this is what you’re looking for: data = str({“state”:1,”endTime”:1518852709307,”fileSize”:000000}) #add a str() here data = data.strip(‘{}’) data = data.split(‘,’) for x in data: x=x.split(‘:’)[-1] # set x to x.split(…) print(x) The script below prints out: 1 1518852709307 0 Here is a one-liner version: print (list(map(lambda … Read more
I’m not sure I understand your “job description”, but I think you want this: def find_matches(tuple_of_dicts, key_to_find): return [d for d in tuple_of_dicts if key_to_find in d] So: >>> tuple_of_dicts = ({18: None}, {10: None}, {16: None, 18: None, 5: None, 6: None, 7: None, 10: None}, {16: None}, {7: None}, {10: None}, {18: None}, … Read more
Gives this: {u’_id’: ObjectId(‘5238273074f8edc6a20c48fe’), u’Command’: u’ABCDEF’, u’processed’: False} But really, all I want is ABCDEF in a string. What you get is a dictionary, you simply need to fetch what you want from it print myDict[ u’Command’] 0 solved pymongo result to String
You are trying to call an unbound function; zxLolBoT is a class, not an instance. As such, the self parameter is not being passed in for you, and the method only received 2 arguments instead of 3 (with self being that 3rd argument). self is already that instance, access the method on that to get … Read more
Supporting Python 3 means supporting Python 3 up to the current point release. Python point releases are backward compatible. What works on 3.0 should generally work on 3.4. See PEP 387 for the general guidelines on this policy. Python 3.4 added some deprecations but none of these will affect packages that were once only written … Read more
Besides from that your function definition for SearchForCapitals has a typo, by which I mean that the line def SerachForCapitals(): should be def SearchForCapitals(): (note the spelling of Search), your code works perfectly fine for me. If this does not fix your problem, you should provide the output you get when running the code. 1 … Read more
How do you do it with zip? I don’t think it get much easier than {chr(96+i):i for i in range(1,27)} Same idea, different indizes, no magic number: {chr(ord(‘a’)+i):i+1 for i in range(26)} 3 solved How to solve this without using zip
This isn’t Python. This is actually C++ code. You didn’t post the error, so I have to assume that you tried to execute this code as Python, which would obviously be your problem. If you called it Python by mistake, and actually got a C++ error, please post the error. 2 solved Invalid Syntax Error … Read more
how about this def display(x): for i in x: for j in i: j = chr(j) print (j, end = ‘ ‘) print() if each i represent a line, you need add a extra print to start over in the next line 1 solved Printing a smile face 🙂
The operators in Python are magic. They invoke special methods depending on the class of the operand. In this case a special method called __mul__ is called from the list class, since a list object is on the left-side. In the case of a list, the list is repeated by the number of times given … Read more
First of all, I am assuming your code is print(“Hello World!”) input(“Press any key to close”) and not print(“Hello World!”) input(“Press any key to close”) To run a python file, all I have to do is py file.py and it works fine. I have also seen people do python file.py on windows, and python3 file.py … Read more
So to answer your questions regarding operators like << or ||= []: << appends an element to an array (or appends strings), in the case above it’s used to append the comment object to either the results array or or the threaded thingy. ||= basically means, if left-hand-side is undefined then evaluate & assign right-hand-side … Read more
This is the Taylor’s series development of exp(-x). Recognizing this gives you a good opportunity to check your result against math.exp(-x). Simple syntax improvements You don’t need an ‘else’ after the while. Just add the code to be run after the loop at the same indentation level as before the while loop. Mathematical problems Most … Read more
Convert each string into a list, then chain the lists: from itertools import chain list(chain.from_iterable(map(list, character))) #[‘#’, ‘$’, ‘@’, ‘#’, ‘%’, ‘%’, ‘#’, ‘@’, ‘#’, ‘%’] 3 solved Adding spaces within the (string) elements of an array [closed]
The first version of dfs calls itself recursively twice, saving the results and reusing them. Here are the two recursive calls: leftpair=dfs(root.left) rightpair=dfs(root.right) The second version of dfs calls itself recursively four times. Here are the four recursive calls, embedded in expressions: left=root.val+dfs(root.left)[1]+dfs(root.right)[1] right=max(dfs(root.left))+max(dfs(root.right)) As you can see, rather than reusing the results of the … Read more