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  • Organizing a random list of objects in Python.

    - by Saebin
    So I have a list that I want to convert to a list that contains a list for each group of objects. ie ['objA.attr1', 'objC', 'objA.attr55', 'objB.attr4'] would return [['objA.attr1', 'objA.attr55'], ['objC'], ['objB.attr4']] currently this is what I use: givenList = ['a.attr1', 'b', 'a.attr55', 'c.attr4'] trgList = [] objNames = [] for val in givenList: obj = val.split('.')[0] if obj in objNames: id = objNames.index(obj) trgList[id].append(val) else: objNames.append(obj) trgList.append([val]) #print trgList It seems to run a decent speed when the original list has around 100,000 ids... but I am curious if there is a better way to do this. Order of the objects or attributes does not matter. Any ideas?

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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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  • Concatenate String to Evernote Markup Language (ENML) in python

    - by Adam the Mediocre
    I am looking to add a string containing the user's text input to the note.content of my note. After reading, I have found how to add resources, but I don't want the resource to be an attachment, I want it to be the actual text. Here is some of the code: title= self.textEditTitle.text() body= self.textEditBody.text() auth_token = "secret stuff!" client = EvernoteClient(token=auth_token, sandbox=True) note_store = client.get_note_store() nBody = "<?xml version=\"1.0\" encoding=\"UTF-8\"?>" nBody += "<!DOCTYPE en-note SYSTEM \"http://xml.evernote.com/pub/enml2.dtd\">" nBody += "<en-note>%s</en-note>" % body note = Types.Note() note.title = title note.content= nBody Any advice would be great, as I'm just starting out with this api and it looks like it's full of potential once I figure it out! Here is what I have been mostly reading from: http://dev.evernote.com/documentation/cloud/chapters/ENML.php

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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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  • 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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  • 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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  • 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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  • 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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  • 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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  • Extract anything that looks like links from large amount of data in python

    - by Riz
    Hi, I have around 5 GB of html data which I want to process to find links to a set of websites and perform some additional filtering. Right now I use simple regexp for each site and iterate over them, searching for matches. In my case links can be outside of "a" tags and be not well formed in many ways(like "\n" in the middle of link) so I try to grab as much "links" as I can and check them later in other scripts(so no BeatifulSoup\lxml\etc). The problem is that my script is pretty slow, so I am thinking about any ways to speed it up. I am writing a set of test to check different approaches, but hope to get some advices :) Right now I am thinking about getting all links without filtering first(maybe using C module or standalone app, which doesn't use regexp but simple search to get start and end of every link) and then using regexp to match ones I need.

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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) 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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  • 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 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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  • 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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