kwargs

Using argparse with function that takes **kwargs argument

由 十年热恋 提交于 2019-12-01 20:47:38
I'm using argparse to take input and pass it to a function that takes as arguments two variables and **kwargs . Here's my function: import requests import sys import argparse def location_by_coordinate(LAT, LNG, **kwargs): if not kwargs: coordinate_url = "https://api.instagram.com/v1/locations/search?lat=%s&lng=%s&access_token=%s" % (LAT, LNG, current_token) r = requests.get(coordinate_url).text else: coordinate_url = "https://api.instagram.com/v1/locations/search?lat=%s&lng=%s&access_token=%s" % (LAT, LNG, current_token) for key, value in kwargs.iteritems(): if 'DISTANCE' in kwargs: distance

Using argparse with function that takes **kwargs argument

由 独自空忆成欢 提交于 2019-12-01 17:49:59
问题 I'm using argparse to take input and pass it to a function that takes as arguments two variables and **kwargs . Here's my function: import requests import sys import argparse def location_by_coordinate(LAT, LNG, **kwargs): if not kwargs: coordinate_url = "https://api.instagram.com/v1/locations/search?lat=%s&lng=%s&access_token=%s" % (LAT, LNG, current_token) r = requests.get(coordinate_url).text else: coordinate_url = "https://api.instagram.com/v1/locations/search?lat=%s&lng=%s&access_token=

Why does the **kwargs mapping compare equal with a differently ordered OrderedDict?

由 倖福魔咒の 提交于 2019-12-01 15:23:10
According to PEP 468 : Starting in version 3.6 Python will preserve the order of keyword arguments as passed to a function. To accomplish this the collected kwargs will now be an ordered mapping . Note that this does not necessarily mean OrderedDict . In that case, why does this ordered mapping fail to respect equality comparison with Python's canonical ordered mapping type, the collections.OrderedDict : >>> from collections import OrderedDict >>> data = OrderedDict(zip('xy', 'xy')) >>> def foo(**kwargs): ... return kwargs == data ... >>> foo(x='x', y='y') # expected result: True True >>> foo

passing functions and its arguments to another function

由 Deadly 提交于 2019-12-01 13:07:37
I have tree types of sub-functions: one without any parameters (arguments), second with one parameter third with multiple parameters (tuple) I am trying to pass that functions and its arguments to another function which sum results of all sub-functions and return the sum value. Parameters in that function should be: names of each sub-function as position arguments (*args) and arguments of each subfunction as key-value arguments (*kvargs). Example : def no_arg() def one_arg(a) def multiple_args(a, b, c, e, f) # execution of function_results_sum: function_results_sum( no_arg, one_arg, multiple

passing functions and its arguments to another function

由 ↘锁芯ラ 提交于 2019-12-01 11:40:06
问题 I have tree types of sub-functions: one without any parameters (arguments), second with one parameter third with multiple parameters (tuple) I am trying to pass that functions and its arguments to another function which sum results of all sub-functions and return the sum value. Parameters in that function should be: names of each sub-function as position arguments (*args) and arguments of each subfunction as key-value arguments (*kvargs). Example : def no_arg() def one_arg(a) def multiple

Python function “remembering” earlier argument (**kwargs)

由 北慕城南 提交于 2019-12-01 07:37:20
I have some objects that have a dictionary of attributes, obj.attrs . The constructor for these objects accepts a dict and/or **kwargs , for convenience. It looks a little like this: class Thing: def __init__(self, attrs={}, **kwargs): for arg in kwargs: attrs[arg] = kwargs[arg] self.attrs = attrs Such that Thing({'color':'red'}) does the same as Thing(color='red') . My problem is that the constructor somehow remembers the last attrs value passed to it. For example: >>> thing1 = Thing(color='red') >>> thing2 = Thing() >>> thing2.attrs {'color': 'red'} ...but thing2.attrs should just be an

Why use packed *args/**kwargs instead of passing list/dict?

由 我只是一个虾纸丫 提交于 2019-11-30 11:37:35
问题 If I don't know how many arguments a function will be passed, I could write the function using argument packing: def add(factor, *nums): """Add numbers and multiply by factor.""" return sum(nums) * factor Alternatively, I could avoid argument packing by passing a list of numbers as the argument: def add(factor, nums): """Add numbers and multiply by factor. :type factor: int :type nums: list of int """ return sum(nums) * factor Is there an advantage to using argument packing *args over passing

passing kwargs with multiprocessing.pool.map

由 一曲冷凌霜 提交于 2019-11-30 11:06:41
问题 I would like to pass keyword arguments to my worker-function with Pool.map(). I can't find a clear example of this when searching forums. Example Code: import multiprocessing as mp def worker((x,y), **kwargs): kwarg_test = kwargs.get('kwarg_test', False) print("kwarg_test = {}".format(kwarg_test)) if kwarg_test: print("Success") return x*y def wrapper_process(**kwargs): jobs = [] pool=mp.Pool(4) for i, n in enumerate(range(4)): jobs.append((n,i)) pool.map(worker, jobs) #works pool.map(worker,

Passing more kwargs into a function than initially set

由 血红的双手。 提交于 2019-11-30 08:30:14
问题 Is there a way to send more kwargs into a function than is called for in the function call? Example: def mydef(a, b): print a print b mydict = {'a' : 'foo', 'b' : 'bar'} mydef(**mydict) # This works and prints 'foo' and 'bar' mybigdict = {'a' : 'foo', 'b' : 'bar', 'c' : 'nooooo!'} mydef(**mybigdict) # This blows up with a unexpected argument error Is there any way to pass in mybigdict without the error? 'c' would never be used in mydef in my ideal world and would just be ignored. Thanks, my

Celery Task Chain and Accessing **kwargs

由 十年热恋 提交于 2019-11-30 07:06:32
I have a situation similar to the one outlined here , except that instead of chaining tasks with multiple arguments, I want to chain tasks that return a dictionary with multiple entries. This is -- very loosely and abstractly --- what I'm trying to do: tasks.py @task() def task1(item1=None, item2=None): item3 = #do some stuff with item1 and item2 to yield item3 return_object = dict(item1=item1, item2=item2, item3=item3) return return_object def task2(item1=None, item2=None, item3=None): item4 = #do something with item1, item2, item3 to yield item4 return_object = dict(item1=item1, item2=item2,
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