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  • smarter "reverse" of a dictionary in python (acc for some of values being the same)?

    - by mrkafk
    def revert_dict(d): rd = {} for key in d: val = d[key] if val in rd: rd[val].append(key) else: rd[val] = [key] return rd >>> revert_dict({'srvc3': '1', 'srvc2': '1', 'srvc1': '2'}) {'1': ['srvc3', 'srvc2'], '2': ['srvc1']} This obviously isn't simple exchange of keys with values: this would overwrite some values (as new keys) which is NOT what I'm after. If 2 or more values are the same for different keys, keys are supposed to be grouped in a list. The above function works, but I wonder if there is a smarter / faster way?

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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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  • Returning a list in this recursive coi function in python.

    - by Nate
    Hello. I'm having trouble getting my list to return in my code. Instead of returning the list, it keeps returning None, but if I replace the return with print in the elif statement, it prints the list just fine. How can I repair this? def makeChange2(amount, coinDenomination, listofcoins = None): #makes a list of coins from an amount given by using a greedy algorithm coinDenomination.sort() #reverse the list to make the largest position 0 at all times coinDenomination.reverse() #assigns list if listofcoins is None: listofcoins = [] if amount >= coinDenomination[0]: listofcoins = listofcoins + [coinDenomination[0]] makeChange2((amount - coinDenomination[0]), coinDenomination, listofcoins) elif amount == 0: return listofcoins else: makeChange2(amount, coinDenomination[1:], listofcoins)

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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 code to do csv file row entries comparison operations and count the number of times row value

    - by Venomancer
    have an excel based CSV file with two columns (or rows, Pythonically) that I am working on. What I need to do is to perform some operations so that I can compare the two data entries in each 'row'. To be more precise, one column has constant numbers all the way down, whereas the other column has varying values. So I need to count the number of times the varying column data entry values crosses the constant value on the other column. For example, fro the csv file i have two columns: Varying Column; Constant Column 24 25 26 25 crossed 27 25 26 25 25.5 25 23 25 crossed 26 25 crossed Thus, the varying column data entries have crossed 25 three times. I need to generate a code that can count the number of the crosses. Please do help out, Thanks.

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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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  • Is this a good approach to execute a list of operations on a data structure in Python?

    - by Sridhar Iyer
    I have a dictionary of data, the key is the file name and the value is another dictionary of its attribute values. Now I'd like to pass this data structure to various functions, each of which runs some test on the attribute and returns True/False. One approach would be to call each function one by one explicitly from the main code. However I can do something like this: #MYmodule.py class Mymodule: def MYfunc1(self): ... def MYfunc2(self): ... #main.py import Mymodule ... #fill the data structure ... #Now call all the functions in Mymodule one by one for funcs in dir(Mymodule): if funcs[:2]=='MY': result=Mymodule.__dict__.get(funcs)(dataStructure) The advantage of this approach is that implementation of main class needn't change when I add more logic/tests to MYmodule. Is this a good way to solve the problem at hand? Are there better alternatives to this solution?

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  • Python : How to add month to December 2012 and get January 2013?

    - by daydreamer
    >>> start_date = date(1983, 11, 23) >>> start_date.replace(month=start_date.month+1) datetime.date(1983, 12, 23) This works until the month is <=11, as soon as I do >>> start_date = date(1983, 12, 23) >>> start_date.replace(month=start_date.month+1) Traceback (most recent call last): File "<stdin>", line 1, in <module> ValueError: month must be in 1..12 How can I keep adding months which increments the year when new month is added to December?

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  • lambda vs. operator.attrGetter('xxx') as sort key 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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  • How to display a page in my browser with python code that is run locally on my computer with "GAE" S

    - by brilliant
    When I run this code on my computer with the help of "Google App Engine SDK", it displays (in my browser) the HTML code of the Google home page: from google.appengine.api import urlfetch url = "http://www.google.com/" result = urlfetch.fetch(url) print result.content How can I make it display the page itself? I mean I want to see that page in my browser the way it would normally be seen by any user of the internet.

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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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  • 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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