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  • Python - Is it possible to get the name of the chained function?

    - by user1326876
    I'm working on a class that basically allows for method chaining, for setting some attrbutes for different dictionaries stored. The syntax is as follows: d = Test() d.connect().setAttrbutes(Message=Blah, Circle=True, Key=True) But there can also be other instances, so, for example: d = Test() d.initialise().setAttrbutes(Message=Blah) Now I believe that I can overwrite the "setattrbutes" function; I just don't want to create a function for each of the dictionary. Instead I want to capture the name of the previous chained function. So in the example above I would then be given "connect" and "initialise" so I know which dictionary to store these inside. I hope this makes sense. Any ideas would be greatly appreciated :)

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  • Add characters (',') every time a certain character ( , )is encountered ? Python 2.7.3

    - by draconisthe0ry
    Let's say you had a string test = 'wow, hello, how, are, you, doing' and you wanted full_list = ['wow','hello','how','are','you','doing'] i know you would start out with an empty list: empty_list = [] and would create a for loop to append the items into a list i'm just confused on how to go about this, I was trying something along the lines of: for i in test: if i == ',': then I get stuck . . .

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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 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 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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  • 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 : 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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  • Best way in Python to determine all possible intersections in a matrix?

    - by ssweens
    So if I have a matrix (list of lists) of unique words as my column headings, document ids as my row headings, and a 0 or 1 as the values if the word exists in that particular document. What I'd like to know is how to determine all the possible combinations of words and documents where more than one word is in common with more than one document. So something like: [[Docid_3, Docid_5], ['word1', 'word17', 'word23']], [[Docid_3, Docid_9, Docid_334], ['word2', 'word7', 'word23', 'word68', 'word982']], and so on for each possible combination. Would love a solution that provides the complete set of combinations and one that yields only the combinations that are not a subset of another, so from the example, not [[Docid_3, Docid_5], ['word1', 'word17']] since it's a complete subset of the first example. I feel like there is an elegant solution that just isn't coming to mind and the beer isn't helping. 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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  • 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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