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  • (Python) Converting a dictionary to a list?

    - by Daria Egelhoff
    So I have this dictionary: ScoreDict = {"Blue": {'R1': 89, 'R2': 80}, "Brown": {'R1': 61, 'R2': 77}, "Purple": {'R1': 60, 'R2': 98}, "Green": {'R1': 74, 'R2': 91}, "Red": {'R1': 87, 'Lon': 74}} Is there any way how I can convert this dictionary into a list like this: ScoreList = [['Blue', 89, 80], ['Brown', 61, 77], ['Purple', 60, 98], ['Green', 74, 91], ['Red', 87, 74]] I'm not too familiar with dictionaries, so I really need some help here. Thanks in advance!

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  • Implement loops for python 3

    - by Alex
    Implement this loop: total up the product of the numbers from 1 to x. Implement this loop: total up the product of the numbers from a to b. Implement this loop: total up the sum of the numbers from a to b. Implement this loop: total up the sum of the numbers from 1 to x. Implement this loop: count the number of characters in a string s. i'm very lost on implementing loops these are just some examples that i am having trouble with-- if someone could help me understand how to do them that would be awesome

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  • filtering elements from list of lists in Python?

    - by user248237
    I want to filter elements from a list of lists, and iterate over the elements of each element using a lambda. For example, given the list: a = [[1,2,3],[4,5,6]] suppose that I want to keep only elements where the sum of the list is greater than N. I tried writing: filter(lambda x, y, z: x + y + z >= N, a) but I get the error: <lambda>() takes exactly 3 arguments (1 given) How can I iterate while assigning values of each element to x, y, and z? Something like zip, but for arbitrarily long lists. thanks, p.s. I know I can write this using: filter(lambda x: sum(x)..., a) but that's not the point, imagine that these were not numbers but arbitrary elements and I wanted to assign their values to variable names.

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  • Dynamically adding @property in python

    - by rz
    I know that I can dynamically add an instance method to an object by doing something like: import types def my_method(self): # logic of method # ... # instance is some instance of some class instance.my_method = types.MethodType(my_method, instance) Later on I can call instance.my_method() and self will be bound correctly and everything works. Now, my question: how to do the exact same thing to obtain the behavior that decorating the new method with @property would give? I would guess something like: instance.my_method = types.MethodType(my_method, instance) instance.my_method = property(instance.my_method) But, doing that instance.my_method returns a property object.

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  • how to use @ in python.. and the @property

    - by zjm1126
    this is my code: def a(): print 'sss' @a() def b(): print 'aaa' b() and the Traceback is: sss Traceback (most recent call last): File "D:\zjm_code\a.py", line 8, in <module> @a() TypeError: 'NoneType' object is not callable so how to use the '@' thanks updated class a: @property def b(x): print 'sss' aa=a() print aa.b it print : sss None how to use @property thanks

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  • Efficient way in Python to remove an element from a comma-separated string

    - by ensnare
    I'm looking for the most efficient way to add an element to a comma-separated string while maintaining alphabetical order for the words: For example: string = 'Apples, Bananas, Grapes, Oranges' subtraction = 'Bananas' result = 'Apples, Grapes, Oranges' Also, a way to do this but while maintaining IDs: string = '1:Apples, 4:Bananas, 6:Grapes, 23:Oranges' subtraction = '4:Bananas' result = '1:Apples, 6:Grapes, 23:Oranges' Sample code is greatly appreciated. Thank you so much.

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  • PYTHON: Look for match in a nested list

    - by elfuego1
    Hello everybody, I have two nested lists of different sizes: A = [[1, 7, 3, 5], [5, 5, 14, 10]] B = [[1, 17, 3, 5], [1487, 34, 14, 74], [1487, 34, 3, 87], [141, 25, 14, 10]] I'd like to gather all nested lists from list B if A[2:4] == B[2:4] and put it into list L: L = [[1, 17, 3, 5], [141, 25, 14, 10]] Would you help me with this?

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  • Python: find <title>

    - by Peter
    I have this: response = urllib2.urlopen(url) html = response.read() begin = html.find('<title>') end = html.find('</title>',begin) title = html[begin+len('<title>'):end].strip() if the url = http://www.google.com then the title have no problem as "Google", but if the url = "http://www.britishcouncil.org/learning-english-gateway" then the title become "<!doctype html public "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML> <HEAD> <base href="http://www.britishcouncil.org/" /> <META http-equiv="Content-Type" Content="text/html;charset=utf-8"> <meta name="WT.sp" content="Learning;Home Page Smart View" /> <meta name="WT.cg_n" content="Learn English Gateway" /> <META NAME="DCS.dcsuri" CONTENT="/learning-english-gateway.htm">..." What is actually happening, why I couldn't return the "title"?

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  • python feedparser with yahoo weather rss

    - by mudder
    I'm trying to use feedparser to get some data from yahoos weather rss. It looks like feed parser strips out the yweather namespace data: http://weather.yahooapis.com/forecastrss?w=24260013&u=c <yweather:condition text="Fair" code="34" temp="23" date="Wed, 19 May 2010 5:55 pm EDT" /> looks like feedparser is completely ignoring that. is there away to get it?

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  • Python: How best to parse a simple grammar?

    - by Rosarch
    Ok, so I've asked a bunch of smaller questions about this project, but I still don't have much confidence in the designs I'm coming up with, so I'm going to ask a question on a broader scale. I am parsing pre-requisite descriptions for a course catalog. The descriptions almost always follow a certain form, which makes me think I can parse most of them. From the text, I would like to generate a graph of course pre-requisite relationships. (That part will be easy, after I have parsed the data.) Some sample inputs and outputs: "CS 2110" => ("CS", 2110) # 0 "CS 2110 and INFO 3300" => [("CS", 2110), ("INFO", 3300)] # 1 "CS 2110, INFO 3300" => [("CS", 2110), ("INFO", 3300)] # 1 "CS 2110, 3300, 3140" => [("CS", 2110), ("CS", 3300), ("CS", 3140)] # 1 "CS 2110 or INFO 3300" => [[("CS", 2110)], [("INFO", 3300)]] # 2 "MATH 2210, 2230, 2310, or 2940" => [[("MATH", 2210), ("MATH", 2230), ("MATH", 2310)], [("MATH", 2940)]] # 3 If the entire description is just a course, it is output directly. If the courses are conjoined ("and"), they are all output in the same list If the course are disjoined ("or"), they are in separate lists Here, we have both "and" and "or". One caveat that makes it easier: it appears that the nesting of "and"/"or" phrases is never greater than as shown in example 3. What is the best way to do this? I started with PLY, but I couldn't figure out how to resolve the reduce/reduce conflicts. The advantage of PLY is that it's easy to manipulate what each parse rule generates: def p_course(p): 'course : DEPT_CODE COURSE_NUMBER' p[0] = (p[1], int(p[2])) With PyParse, it's less clear how to modify the output of parseString(). I was considering building upon @Alex Martelli's idea of keeping state in an object and building up the output from that, but I'm not sure exactly how that is best done. def addCourse(self, str, location, tokens): self.result.append((tokens[0][0], tokens[0][1])) def makeCourseList(self, str, location, tokens): dept = tokens[0][0] new_tokens = [(dept, tokens[0][1])] new_tokens.extend((dept, tok) for tok in tokens[1:]) self.result.append(new_tokens) For instance, to handle "or" cases: def __init__(self): self.result = [] # ... self.statement = (course_data + Optional(OR_CONJ + course_data)).setParseAction(self.disjunctionCourses) def disjunctionCourses(self, str, location, tokens): if len(tokens) == 1: return tokens print "disjunction tokens: %s" % tokens How does disjunctionCourses() know which smaller phrases to disjoin? All it gets is tokens, but what's been parsed so far is stored in result, so how can the function tell which data in result corresponds to which elements of token? I guess I could search through the tokens, then find an element of result with the same data, but that feel convoluted... What's a better way to approach this problem?

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  • Python continue from the point where exception was thrown

    - by James Lin
    Hi is there a way to continue from the point where exception was thrown? eg I have the following psudo code unique code 1 unique code 2 unique code 3 if I want to ignore the exceptions of any of the unique code statements I will have to do it like this: try: #unique code 1 except: pass try: #unique code 2 except: pass try: #unique code 3 except: pass but this isn't elegant to me, and for the life of me I can't remember how I resolved this kind of problem last time... what I want to have is something like try: unique code 1 unique code 2 unique code 3 except: continue from last exception raised

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  • Custom keys for Google App Engine models (Python)

    - by Cameron
    First off, I'm relatively new to Google App Engine, so I'm probably doing something silly. Say I've got a model Foo: class Foo(db.Model): name = db.StringProperty() I want to use name as a unique key for every Foo object. How is this done? When I want to get a specific Foo object, I currently query the datastore for all Foo objects with the target unique name, but queries are slow (plus it's a pain to ensure that name is unique when each new Foo is created). There's got to be a better way to do this! Thanks.

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  • has any simply way to delete a value in list of python

    - by zjm1126
    a=[1,2,3,4] b=a.index(6) del a[b] print a it show error: Traceback (most recent call last): File "D:\zjm_code\a.py", line 6, in <module> b=a.index(6) ValueError: list.index(x): x not in list so i have to do this : a=[1,2,3,4] try: b=a.index(6) del a[b] except: pass print a but this is not simple,has any simply way ? thanks

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  • Unique elements of list within list in python

    - by user2901061
    We are given a list of animals in different zoos and need to find which zoos have animals that are not in any others. The animals of each zoo are separated by spaces, and each zoo is originally separated by a comma. I am currently enumerating over all of the zoos to split each animal and create lists within lists for different zoos as such: for i, zoo in enumerate(zoos): zoos[i] = zoo.split() However, I then do not know how to tell and count how many of the zoos have unique animals. I figure it is something else with enumerate and possibly sets, but cannot get it down exactly. Any help is greatly appreciated. Thanks

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  • How to break the following line of python

    - by FrederikNS
    Hello Stack Overflow, I have come upon a couple of lines of code similar to this one, but I'm unsure how I should break it: blueprint = Blueprint(self.blueprint_map[str(self.ui.blueprint_combo.currentText())], runs=self.ui.runs_spin.text(), me=self.ui.me_spin.text(), pe=self.ui.pe_skill_combo.currentIndex()) Thanks in advance

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  • os.walk in python not running with cmd line parameter passed as path

    - by kartiku
    Hello, I needed to find the number of files in a folder on the system. This is what i used: file_count = sum((len(f) for _, _, f in os.walk('path'))) This works fine when we specify the path as a string in quotes, but when I enter a variable name that holds the path, type(file_count) is a generator object, and hence cannot be used as an integer. How to solve this and why does this happen?

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  • Python combinations no repeat by constraint

    - by user2758113
    I have a tuple of tuples (Name, val 1, val 2, Class) tuple = (("Jackson",10,12,"A"), ("Ryan",10,20,"A"), ("Michael",10,12,"B"), ("Andrew",10,20,"B"), ("McKensie",10,12,"C"), ("Alex",10,20,"D")) I need to return all combinations using itertools combinations that do not repeat classes. How can I return combinations that dont repeat classes. For example, the first returned statement would be: tuple0, tuple2, tuple4, tuple5 and so on.

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  • Using Memcached in Python/Django - questions.

    - by Thomas
    I am starting use Memcached to make my website faster. For constant data in my database I use this: from django.core.cache import cache cache_key = 'regions' regions = cache.get(cache_key) if result is None: """Not Found in Cache""" regions = Regions.objects.all() cache.set(cache_key, regions, 2592000) #(2592000sekund = 30 dni) return regions For seldom changed data I use signals: from django.core.cache import cache from django.db.models import signals def nuke_social_network_cache(self, instance, **kwargs): cache_key = 'networks_for_%s' % (self.instance.user_id,) cache.delete(cache_key) signals.post_save.connect(nuke_social_network_cache, sender=SocialNetworkProfile) signals.post_delete.connect(nuke_social_network_cache, sender=SocialNetworkProfile) Is it correct way? I installed django-memcached-0.1.2, which show me: Memcached Server Stats Server Keys Hits Gets Hit_Rate Traffic_In Traffic_Out Usage Uptime 127.0.0.1 15 220 276 79% 83.1 KB 364.1 KB 18.4 KB 22:21:25 Can sombody explain what columns means? And last question. I have templates where I am getting much records from a few table (relationships). So in my view I get records from one table and in templates show it and related info from others. Generating page last a few seconds for very small table (<100records). Is it some easy way to cache queries from templates? Have I to do some big structure in my view (with all related tables), cache it and send to template?

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  • Python function argument scope (Dictionaries v. Strings)

    - by Shaun Meyer
    Hello, given: foo = "foo" def bar(foo): foo = "bar" bar(foo) print foo # foo is still "foo"... foo = {'foo':"foo"} def bar(foo): foo['foo'] = "bar" bar(foo) print foo['foo'] # foo['foo'] is now "bar"? I have a function that has been inadvertently over-writing my function parameters when I pass a dictionary. Is there a clean way to declare my parameters as constant or am I stuck making a copy of the dictionary within the function? Thanks!

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  • Join a list of lists together into 1 list in Python

    - by dotty
    Hay All. I have a list which consists of many lists, here is an example [ [Obj, Obj, Obj, Obj], [Obj], [Obj], [ [Obj,Obj], [Obj,Obj,Obj] ] ] Is there a way to join all these items together as 1 list, so the output will be something like [Obj,Obj,Obj,Obj,Obj,Obj,Obj,Obj,Obj,Obj,Obj] Thanks

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