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  • How to test that variable is not equal to multiple things? Python

    - by M830078h
    This is the piece of code I have: choice = "" while choice != "1" and choice != "2" and choice != "3": choice = raw_input("pick 1, 2 or 3") if choice == "1": print "1 it is!" elif choice == "2": print "2 it is!" elif choice == "3": print "3 it is!" else: print "You should choose 1, 2 or 3" While it works, I feel that it's really clumsy, specifically the while clause. What if I have more acceptable choices? Is there a better way to make the clause?

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  • How to add values accordingly of the first indices of a dictionary of tuples of a list of strings? Python 3x

    - by TheStruggler
    I'm stuck on how to formulate this problem properly and the following is: What if we had the following values: {('A','B','C','D'):3, ('A','C','B','D'):2, ('B','D','C','A'):4, ('D','C','B','A'):3, ('C','B','A','D'):1, ('C','D','A','B'):1} When we sum up the first place values: [5,4,2,3] (5 people picked for A first, 4 people picked for B first, and so on like A = 5, B = 4, C = 2, D = 3) The maximum values for any alphabet is 5, which isn't a majority (5/14 is less than half), where 14 is the sum of total values. So we remove the alphabet with the fewest first place picks. Which in this case is C. I want to return a dictionary where {'A':5, 'B':4, 'C':2, 'D':3} without importing anything. This is my work: def popular(letter): '''(dict of {tuple of (str, str, str, str): int}) -> dict of {str:int} ''' my_dictionary = {} counter = 0 for (alphabet, picks) in letter.items(): if (alphabet[0]): my_dictionary[alphabet[0]] = picks else: my_dictionary[alphabet[0]] = counter return my_dictionary This returns duplicate of keys which I cannot get rid of. Thanks.

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  • (Python/Pyramid) Better ways to have standard list/form editors?

    - by badcat
    I'm working on a number of Pyramid (former Pylons) projects, and often I have the need to display a list of some content (let's say user accounts, log entries or simply some other data). A user should be able to paginate through the list, click on a row and get a form where he/she can edit the contents of that row. Right now I'm always re-inventing the wheel by having Mako templates which use webhelpers for the pagination, Jquery UI for providing a dialog and I craft the editor form and AJAX requests on the client and server side by hand. As you may know, this eats up painfully much time. So what I'm wondering is: Is there a better way of providing lists, editor dialog and server/client communication about this, without having to re-invent the wheel every time? I heard Django takes off a big load of that by providing user accounts and other stuff out of the box; but in my case it's not just about user accounts, it can be any kind of data that is stored on the server-side in a SQL database, which should be able to be edited by a user. Thanks in advance!

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  • How to remove certain lists from a list of lists using python?

    - by seaworthy
    I can not figure out why my code does not filter out lists from a predefined list. I am trying to remove specific list using the following code. data = [[1,1,1],[1,1,2],[1,2,1],[1,2,2],[2,1,1],[2,1,2],[2,2,1],[2,2,2]] data = [x for x in data if x[0] != 1 and x[1] != 1] print data My result: data = [[2, 2, 1], [2, 2, 2]] Expected result: data = [[1,2,1],[1,2,2],[2,1,1],[2,1,2],[2,2,1],[2,2,2]]

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  • Python - Find where in string regex match was found?

    - by nb
    I'm currently using regular expressions to search through RSS feeds to find if certain words and phrases are mentioned, and would then like to extract the text on either side of the match as well. For example: String = "This is an example sentence, it is for demonstration only" re.search("is", String) I'd like to know where the is was found so that I can extract and output something like this: 1 match found: "This is an example sentence" I know that it would be easy to do with splits, but I'd need to know what the index of first character of the match was in the string, which I don't know how to find

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  • The return value should be a list but doesn't return as expected?! - Python newbie

    - by user1432941
    Hi this must be a very simple solution that has eluded me this last hour. I've tried to build this test function where the return value of the test_cases list should match the values in the test_case_answers list but for some reason, test case 1 and test case 2 fail. When i print the return values for the test cases they return the correct answers, but for some reason test case 1 and test case 2 return False. Thanks for your help! import math test_cases = [1, 9, -3] test_case_answers = [1, 3, 0] def custom_sqrt(num): for i in range(len(test_cases)): if test_cases[i] >= 0: return math.sqrt(test_cases[i]) else: return 0 for i in range(len(test_cases)): if custom_sqrt(test_cases[i]) != test_case_answers[i]: print "Test Case #", i, "failed!" custom_sqrt(test_cases)

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  • Python raises a KeyError (for an out of dictionary key) even though the key IS in the dictionary

    - by ignorantslut
    I'm getting a KeyError for an out of dictionary key, even though I know the key IS in fact in the dictionary. Any ideas as to what might be causing this? print G.keys() returns the following: ['24', '25', '20', '21', '22', '23', '1', '3', '2', '5', '4', '7', '6', '9', '8', '11', '10', '13', '12', '15', '14', '17', '16', '19', '18'] but when I try to access a value in the dictionary on the next line of code... for w in G[v]: #note that in this example, v = 17 I get the following error message: KeyError: 17 Any help, tips, or advice are all appreciated. Thanks.

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  • Are python list comprehensions always a good programming practice?

    - by dln385
    To make the question clear, I'll use a specific example. I have a list of college courses, and each course has a few fields (all of which are strings). The user gives me a string of search terms, and I return a list of courses that match all of the search terms. This can be done in a single list comprehension or a few nested for loops. Here's the implementation. First, the Course class: class Course: def __init__(self, date, title, instructor, ID, description, instructorDescription, *args): self.date = date self.title = title self.instructor = instructor self.ID = ID self.description = description self.instructorDescription = instructorDescription self.misc = args Every field is a string, except misc, which is a list of strings. Here's the search as a single list comprehension. courses is the list of courses, and query is the string of search terms, for example "history project". def searchCourses(courses, query): terms = query.lower().strip().split() return tuple(course for course in courses if all( term in course.date.lower() or term in course.title.lower() or term in course.instructor.lower() or term in course.ID.lower() or term in course.description.lower() or term in course.instructorDescription.lower() or any(term in item.lower() for item in course.misc) for term in terms)) You'll notice that a complex list comprehension is difficult to read. I implemented the same logic as nested for loops, and created this alternative: def searchCourses2(courses, query): terms = query.lower().strip().split() results = [] for course in courses: for term in terms: if (term in course.date.lower() or term in course.title.lower() or term in course.instructor.lower() or term in course.ID.lower() or term in course.description.lower() or term in course.instructorDescription.lower()): break for item in course.misc: if term in item.lower(): break else: continue break else: continue results.append(course) return tuple(results) That logic can be hard to follow too. I have verified that both methods return the correct results. Both methods are nearly equivalent in speed, except in some cases. I ran some tests with timeit, and found that the former is three times faster when the user searches for multiple uncommon terms, while the latter is three times faster when the user searches for multiple common terms. Still, this is not a big enough difference to make me worry. So my question is this: which is better? Are list comprehensions always the way to go, or should complicated statements be handled with nested for loops? Or is there a better solution altogether?

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  • I'm making a resume...what would you consider as intermediate python programming skills?

    - by user285884
    I've made a couple of scripts. One is a stock screener that can search through every stock. Another creates a heatmap that tells you what's performed well and badly over the past day. They aren't really that useful, just did them to work on my programming skills. I was able to throw some SQL in my scripts too. Would you call that intermediate? Thanks? How do you guys list your programming skills on your resume? Maybe there's a better way of putting it on my resume than "intermediate" or "beginner."

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  • How to select random image of specific size using Django / Python?

    - by Jonathan
    I've been using this little snippet to select random images. However I would like to change it to select only images of a certain size. I'm running into trouble checking against image size. If I use get_image_dimensions() I need to use a conditional statement, which then requires that I allow exceptions. So, I guess I need some pointers on just limiting by image dimensions. Thanks. import os import random import posixpath from django import template from django.conf import settings register = template.Library() def is_image_file(filename): """Does `filename` appear to be an image file?""" img_types = [".jpg", ".jpeg", ".png", ".gif"] ext = os.path.splitext(filename)[1] return ext in img_types @register.simple_tag def random_image(path): """ Select a random image file from the provided directory and return its href. `path` should be relative to MEDIA_ROOT. Usage: <img src='{% random_image "images/whatever/" %}'> """ fullpath = os.path.join(settings.MEDIA_ROOT, path) filenames = [f for f in os.listdir(fullpath) if is_image_file(f)] pick = random.choice(filenames) return posixpath.join(settings.MEDIA_URL, path, pick)

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  • What is the difference between "a is b" and "id(a) == id(b)" in Python?

    - by bp
    The id() inbuilt function gives... an integer (or long integer) which is guaranteed to be unique and constant for this object during its lifetime. The is operator, instead, gives... object identity So why is it possible to have two objects that have the same id but return False to an is check? Here is an example: >>> class Test(): ... def test(): ... pass >>> a = Test() >>> b = Test() >>> id(a.test) == id(b.test) True >>> a.test is b.test False A more troubling example: (continuing the above) >>> b = a >>> b is a True >>> b.test is a.test False >>> a.test is a.test False

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  • lambda vs. operator.attrgetter('xxx') as sort key function in Python

    - by Paul McGuire
    I am looking at some code that has a lot of sort calls using comparison functions, and it seems like it should be using key functions. If you were to change seq.sort(lambda x,y: cmp(x.xxx, y.xxx)), which is preferable: seq.sort(key=operator.attrgetter('xxx')) or: seq.sort(key=lambda a:a.xxx) I would also be interested in comments on the merits of making changes to existing code that works.

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  • Python - Is a dictionary slow to find frequency of each character?

    - by psihodelia
    I am trying to find a frequency of each symbol in any given text using an algorithm of O(n) complexity. My algorithm looks like: s = len(text) P = 1.0/s freqs = {} for char in text: try: freqs[char]+=P except: freqs[char]=P but I doubt that this dictionary-method is fast enough, because it depends on the underlying implementation of the dictionary methods. Is this the fastest method?

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