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  • Python 3: Most efficient way to create a [func(i) for i in range(N)] list comprehension

    - by mejiwa
    Say I have a function func(i) that creates an object for an integer i, and N is some nonnegative integer. Then what's the fastest way to create a list (not a range) equal to this list mylist = [func(i) for i in range(N)] without resorting to advanced methods like creating a function in C? My main concern with the above list comprehension is that I'm not sure if python knows beforehand the length of range(N) to preallocate mylist, and therefore has to incrementally reallocate the list. Is that the case or is python clever enough to allocate mylist to length N first and then compute it's elements? If not, what's the best way to create mylist? Maybe this? mylist = [None]*N for i in range(N): mylist[i] = func(i)

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  • Counting entries in a list of dictionaries: for loop vs. list comprehension with map(itemgetter)

    - by Dennis Williamson
    In a Python program I'm writing I've compared using a for loop and increment variables versus list comprehension with map(itemgetter) and len() when counting entries in dictionaries which are in a list. It takes the same time using a each method. Am I doing something wrong or is there a better approach? Here is a greatly simplified and shortened data structure: list = [ {'key1': True, 'dontcare': False, 'ignoreme': False, 'key2': True, 'filenotfound': 'biscuits and gravy'}, {'key1': False, 'dontcare': False, 'ignoreme': False, 'key2': True, 'filenotfound': 'peaches and cream'}, {'key1': True, 'dontcare': False, 'ignoreme': False, 'key2': False, 'filenotfound': 'Abbott and Costello'}, {'key1': False, 'dontcare': False, 'ignoreme': True, 'key2': False, 'filenotfound': 'over and under'}, {'key1': True, 'dontcare': True, 'ignoreme': False, 'key2': True, 'filenotfound': 'Scotch and... well... neat, thanks'} ] Here is the for loop version: #!/usr/bin/env python # Python 2.6 # count the entries where key1 is True # keep a separate count for the subset that also have key2 True key1 = key2 = 0 for dictionary in list: if dictionary["key1"]: key1 += 1 if dictionary["key2"]: key2 += 1 print "Counts: key1: " + str(key1) + ", subset key2: " + str(key2) Output for the data above: Counts: key1: 3, subset key2: 2 Here is the other, perhaps more Pythonic, version: #!/usr/bin/env python # Python 2.6 # count the entries where key1 is True # keep a separate count for the subset that also have key2 True from operator import itemgetter KEY1 = 0 KEY2 = 1 getentries = itemgetter("key1", "key2") entries = map(getentries, list) key1 = len([x for x in entries if x[KEY1]]) key2 = len([x for x in entries if x[KEY1] and x[KEY2]]) print "Counts: key1: " + str(key1) + ", subset key2: " + str(key2) Output for the data above (same as before): Counts: key1: 3, subset key2: 2 I'm a tiny bit surprised these take the same amount of time. I wonder if there's something faster. I'm sure I'm overlooking something simple. One alternative I've considered is loading the data into a database and doing SQL queries, but the data doesn't need to persist and I'd have to profile the overhead of the data transfer, etc., and a database may not always be available. I have no control over the original form of the data. The code above is not going for style points.

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  • convert string to dict using list comprehension in python

    - by Pavel
    I have came across this problem a few times and can't seem to figure out a simple solution. Say I have a string string = "a=0 b=1 c=3" I want to convert that into a dictionary with a, b and c being the key and 0, 1, and 3 being their respective values (converted to int). Obviously I can do this: list = string.split() dic = {} for entry in list: key, val = entry.split('=') dic[key] = int(val) But I don't really like that for loop, It seems so simple that you should be able to convert it to some sort of list comprehension expression. And that works for slightly simpler cases where the val can be a string. dic = dict([entry.split('=') for entry in list]) However, I need to convert val to an int on the fly and doing something like this is syntactically incorrect. dic = dict([[entry[0], int(entry[1])] for entry.split('=') in list]) So my question is: is there a way to eliminate the for loop using list comprehension? If not, is there some built in python method that will do that for me?

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  • nested list comprehension using intermediate result

    - by KentH
    I am trying to grok the output of a function which doesn't have the courtesy of setting a result code. I can tell it failed by the "error:" string which is mixed into the stderr stream, often in the middle of a different conversion status message. I have the following list comprehension which works, but scans for the "error:" string twice. Since it is only rescanning the actual error lines, it works fine, but it annoys me I can't figure out how to use a single scan. Here's the working code: errors = [e[e.find('error:'):] for e in err.splitlines() if 'error:' in e] The obvious (and wrong) way to simplify is to save the "find" result errors = [e[i:] for i in e.find('error:') if i != -1 for e in err.splitlines()] However, I get "UnboundLocalError: local variable 'e' referenced before assignment". Blindly reversing the 'for's in the comprehension also fails. How is this done? THanks. Kent

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  • How do I make this simple list comprehension?

    - by Carson Myers
    I'm new to python, and I'm trying to get to know the list comprehensions better. I'm not even really sure if list comprehension is the word I'm looking for, since I'm not generating a list. But I am doing something similar. This is what I am trying to do: I have a list of numbers, the length of which is divisible by three. So say I have nums = [1, 2, 3, 4, 5, 6] I want to iterate over the list and get the sum of each group of three digits. Currently I am doing this: for i in range(0, len(nums), 3): nsum = a + b + c for a, b, c in nums[i, i+3] print(nsum) I know this is wrong, but is there a way to do this? I'm sure I've overlooked something probably very simple... But I can't think of another way to do this.

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  • How do you solve this Haskell problem?

    - by Linda Cohen
    I want to define a function replicate to replicate a list of numbers by its value using only list comprehension, for example: replicate [5,1,3,2,8,1,2] output: [5,5,5,5,5,1,3,3,3,2,2,8,8,8,8,8,8,8,8,1,2,2] I know this would be easy to use the 'replicate' built in function but only list comprehension is allow, how can I do this? THANKS!

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  • List filtering: list comprehension vs. lambda + filter

    - by Agos
    I happened to find myself having a basic filtering need: I have a list and I have to filter it by an attribute of the items. My code looked like this: list = [i for i in list if i.attribute == value] But then i thought, wouldn't it be better to write it like this? filter(lambda x: x.attribute == value, list) It's more readable, and if needed for performance the lambda could be taken out to gain something. Question is: are there any caveats in using the second way? Any performance difference? Am I missing the Pythonic Way™ entirely and should do it in yet another way (such as using itemgetter instead of the lambda)? Thanks in advance

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  • What is wrong with this list comprehension code?

    - by suresh
    My aim is to list all elements of the array a whose values are greater than their index positions. I wrote a Haskell code like this. [a|i<-[0..2],a<-[1..3],a!!i>i] When tested on ghci prelude prompt, I get the following error message which I am unable to understand. No instance for (Num [a]) arising from the literal 3 at <interactive>:1:20 Possible fix: add an instance declaration for (Num [a])

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  • List comprehension, map, and numpy.vectorize performance

    - by mcstrother
    I have a function foo(i) that takes an integer and takes a significant amount of time to execute. Will there be a significant performance difference between any of the following ways of initializing a: a = [foo(i) for i in xrange(100)] a = map(foo, range(100)) vfoo = numpy.vectorize(foo) a = vfoo(range(100)) (I don't care whether the output is a list or a numpy array.) Is there a better way?

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  • Understanding Scope on Scala's For Loops (For Comprehension)

    - by T. Stone
    In Chapter 3 of Programming Scala, the author gives two examples of for loops / for comprehensions, but switches between using ()'s and {}'s. Why is this the case, as these inherently look like they're doing the same thing? Is there a reason breed <- dogBreeds is on the 2nd line in example #2? // #1 ()'s for (breed <- dogBreeds if breed.contains("Terrier"); if !breed.startsWith("Yorkshire") ) println(breed) // #2 {}'s for { breed <- dogBreeds upcasedBreed = breed.toUpperCase() } println(upcasedBreed)

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  • Efficiently generate numpy array from list comprehension output?

    - by shootingstars
    Is there a more efficient way than using numpy.asarray() to generate an array from output in the form of a list? This appears to be copying everything in memory, which doesn't seem like it would be that efficient with very large arrays. (Updated) Example: import numpy as np a1 = np.array([1,2,3,4,5,6,7,8,9,10]) # pretend this has thousands of elements a2 = np.array([3,7,8]) results = np.asarray([np.amax(np.where(a1 > element)) for element in a2])

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  • List comprehension in Ruby

    - by Readonly
    To do the equivalent of Python list comprehensions, I'm doing the following: some_array.select{|x| x % 2 == 0 }.collect{|x| x * 3} Is there a better way to do this...perhaps with one method call?

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  • Python - List of Lists Slicing Behavior

    - by Dan Dobint
    When I define a list and try to change a single item like this: list_of_lists = [['a', 'a', 'a'], ['a', 'a', 'a'], ['a', 'a', 'a']] list_of_lists[1][1] = 'b' for row in list_of_lists: print row It works as intended. But when I try to use list comprehension to create the list: row = ['a' for range in xrange(3)] list_of_lists = [row for range in xrange(3)] list_of_lists[1][1] = 'b' for row in list_of_lists: print row It results in an entire column of items in the list being changed. Why is this? How can I achieve the desired effect with list comprehension?

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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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  • Excel CSV into Nested Dictionary; List Comprehensions

    - by victorhooi
    heya, I have a Excel CSV files with employee records in them. Something like this: mail,first_name,surname,employee_id,manager_id,telephone_number [email protected],john,smith,503422,503423,+65(2)3423-2433 [email protected],george,brown,503097,503098,+65(2)3423-9782 .... I'm using DictReader to put this into a nested dictionary: import csv gd_extract = csv.DictReader(open('filename 20100331 original.csv'), dialect='excel') employees = dict([(row['employee_id'], row) for row in gp_extract]) Is the above the proper way to do it - it does work, but is it the Right Way? Something more efficient? Also, the funny thing is, in IDLE, if I try to print out "employees" at the shell, it seems to cause IDLE to crash (there's approximately 1051 rows). 2. Remove employee_id from inner dict The second issue issue, I'm putting it into a dictionary indexed by employee_id, with the value as a nested dictionary of all the values - however, employee_id is also a key:value inside the nested dictionary, which is a bit redundant? Is there any way to exclude it from the inner dictionary? 3. Manipulate data in comprehension Thirdly, we need do some manipulations to the imported data - for example, all the phone numbers are in the wrong format, so we need to do some regex there. Also, we need to convert manager_id to an actual manager's name, and their email address. Most managers are in the same file, while others are in an external_contractors CSV, which is similar but not quite the same format - I can import that to a separate dict though. Are these two items things that can be done within the single list comprehension, or should I use a for loop? Or does multiple comprehensions work? (sample code would be really awesome here). Or is there a smarter way in Python do it? Cheers, Victor

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  • Ouverture de la rubrique Cloud Computing, pour trouver les ressources nécessaires à la compréhension et à l'utilisation du "Cloud"

    Bonjour à tous, La rubrique Cloud Computing vient de voir le jour à l'adresse http://cloud-computing.developpez.com. Cette rubrique contiendra des news et toutes les ressources nécessaires à la compréhension, à l'utilisation et au développement pour et avec le "Cloud". Si vous avez des idées de tutoriels, d'articles, de sources ou encore de Q/R pour de prochaines FAQ, n'hésitez pas à nous en faire part. Très cordialement, Gordon...

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  • Ouverture de la rubrique Cloud Computing, pour trouver les ressources nécessaires à la compréhension et à l'utilisation du "Cloud"

    Bonjour à tous, La rubrique Cloud Computing vient de voir le jour à l'adresse http://cloud-computing.developpez.com. Cette rubrique contiendra des news et toutes les ressources nécessaires à la compréhension, à l'utilisation et au développement pour et avec le "Cloud" (de Windows Azure aux Google Apps en passant par Salesforce et les serveurs HPC). Si vous avez des idées de tutoriels, d'articles, de sources ou encore de Q/R pour de prochaines FAQ, n'hésitez pas à nous en faire part. Très cordialement, Gordon...

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  • Python - How to find a correlation between two vectors ?

    - by psihodelia
    Given two vectors X and Y I have to find their correlation, i.e. their linear dependence/independence. Both vectors have equal dimension. A resulted answer should be a floating point number from [-1.0 .. 1.0]. Example: X=[-1, 2, 0] Y=[ 4, 2, -0.3] Find y=cor(X,Y) such that y belongs to [-1.0 .. 1.0]. It should be a simple construction involving a list-comprehension. No external library is allowed. UPDATE: ok, if dot product is enough, then here is my solution: nX = 1/(sum([x*x for x in X]) ** 0.5) nY = 1/(sum([y*y for y in Y]) ** 0.5) cor = sum([(x*nX)*(y*nY) for x,y in zip(X,Y) ]) right?

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  • Show me some cool python list comprehensions

    - by christangrant
    One of the major strengths of python and a few other (functional) programming languages are the list comprehension. They allow programmers to write complex expressions in 1 line. They may be confusing at first but if one gets used to the syntax, it is much better than nested complicated for loops. With that said, please share with me some of the coolest uses of list comprehensions. (By cool, I just mean useful) It could be for some programming contest, or a production system. For example: To do the transpose of a matrix mat >>> mat = [ ... [1, 2, 3], ... [4, 5, 6], ... [7, 8, 9], ... ] >>> [[row[i] for row in mat] for i in [0, 1, 2]] [[1, 4, 7], [2, 5, 8], [3, 6, 9]] Please include a description of the expression and where it was used (if possible).

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  • Update a list from another list

    - by Langali
    I have a list of users in local store that I need to update from a remote list of users every once in a while. Basically: If a remote user already exists locally, update its fields. If a remote user doesn't already exist locally, add the user. If a local user doesn't appear in the remote list, deactivate or delete. If a local user also appears in the remote list, update its fields. Just a simple case of syncing the local list. Is there a better way to do this in pure Java than the following? I feel gross looking at my own code. public class User { Integer id; String email; boolean active; //Getters and Setters....... public User(Integer id, String email, boolean active) { this.id = id; this.email = email; this.active = active; } @Override public boolean equals(Object other) { boolean result = false; if (other instanceof User) { User that = (User) other; result = (this.getId() == that.getId()); } return result; } } public static void main(String[] args) { //From 3rd party List<User> remoteUsers = getRemoteUsers(); //From Local store List<User> localUsers =getLocalUsers(); for (User remoteUser : remoteUsers) { boolean found = false; for (User localUser : localUsers) { if (remoteUser.equals(localUser)) { found = true; localUser.setActive(remoteUser.isActive()); localUser.setEmail(remoteUser.getEmail()); //update } break; } if (!found) { User user = new User(remoteUser.getId(), remoteUser.getEmail(), remoteUser.isActive()); //Save } } for(User localUser : localUsers ) { boolean found = false; for(User remoteUser : remoteUsers) { if(localUser.equals(remoteUser)) { found = true; localUser.setActive(remoteUser.isActive()); localUser.setEmail(remoteUser.getEmail()); //Update } break; } if(!found) { localUser.setActive(false); // Deactivate } } }

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  • Haskell Lazy Evaluation and Reuse

    - by Jonathan Sternberg
    I know that if I were to compute a list of squares in Haskell, I could do this: squares = [ x ** 2 | x <- [1 ..] ] Then when I call squares like this: print $ take 4 squares And it would print out [1.0, 4.0, 9.0, 16.0]. This gets evaluated as [ 1 ** 2, 2 ** 2, 3 ** 2, 4 ** 2 ]. Now since Haskell is functional and the result would be the same each time, if I were to call squares again somewhere else, would it re-evaluate the answers it's already computed? If I were to re-use squares after I had already called the previous line, would it re-calculate the first 4 values? print $ take 5 squares Would it evaluate [1.0, 4.0, 9.0, 16.0, 5 ** 2]?

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