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Lazy evaluation of a Python dictionary


Lazy evaluation of a Python dictionary

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Lazy evaluation of a Python dictionary
Tag : python , By : Schmidt
Date : November 25 2020, 03:01 PM

To fix the issue you can do You could use a modified version (below) of the run_once decorator described in this answer.
def run_once(f):
    def wrapper(*args, **kwargs):
        if not wrapper.has_run:
             wrapper.has_run = True
             wrapper.cache = f(*args, **kwargs)
        return wrapper.cache
    wrapper.has_run = False
    return wrapper

@run_once
def getMyDict():
    myDict = {}
    myDict['1'] = '1'
    print('Run getMyDict')
    return myDict


print(getMyDict()['1'])
print(getMyDict()['1'])
print(getMyDict()['1'])
Run getMyDict
1
1
1
def initMyDict():
    myDict = {}
    myDict['1'] = '1'
    print('Run initMyDict')
    return myDict
def getMyDict(myDict=initMyDict()):
    return myDict
print(getMyDict()['1'])
print(getMyDict()['1'])
print(getMyDict()['1'])
Run initMyDict
1
1
1

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Lazy evaluation in Python? Between modules?


Tag : python , By : CM.
Date : March 29 2020, 07:55 AM
To fix the issue you can do Python is dynamically typed, so this is nothing fancy:
configMod.py:
##### this is configMod.py #####
modTestArg1 = "Brian"

def runTest(testMod, *args, **kwargs):
    testMod.doTest(*args, **kwargs)
##### this is testMod.py #####
def doTest(name, address, city="San Francisco"):
    print "doTest called with name: %s, address: %s, city: %s" % \
        (name, address, city)
##### this is controlMod.py #####
import configMod
import testMod

configMod.runTest(testMod, configMod.modTestArg1, "123 Fake Street")
python controlMod.py
doTest called with name: Brian, address: 123 Fake Street, city: San Francisco

Lazy evaluation in Python


Tag : python , By : user87752
Date : March 29 2020, 07:55 AM
will be helpful for those in need The object returned by range() (or xrange() in Python2.x) is known as a lazy iterable.
Instead of storing the entire range, [0,1,2,..,9], in memory, the generator stores a definition for (i=0; i<10; i+=1) and computes the next value only when needed (AKA lazy-evaluation).
# this is a list, create all 5000000 x/2 values immediately, uses []
lis = [x/2 for x in range(5000000)]

# this is a generator, creates each x/2 value only when it is needed, uses ()
gen = (x/2 for x in range(5000000)) 
# this is also a generator, it will run until a yield occurs, and return that result.
# on the next call it picks up where it left off and continues until a yield occurs...
def divby2(n):
    num = 0
    while num < n:
        yield num/2
        num += 1

# same as (x/2 for x in range(5000000))
print divby2(5000000)

Python lazy evaluation?


Tag : python , By : Juan Pablo
Date : March 29 2020, 07:55 AM
this will help The misconception here is that this statement is not being immediately invoked:
is_skip_line = self.is_skip_line(input_line)

Delay an evaluation / lazy evaluation in Python


Tag : python , By : CodeOfficer
Date : March 29 2020, 07:55 AM
this one helps. a simple lambda works. When called, the function will fetch test_class variable from the current scope, and if it finds it, that will work, like below:
delayed_evaluation_0 = lambda : test_class.get_variable_0()
test_class = TestClass(3)
print(delayed_evaluation_0())

Python lazy dictionary evaluation


Tag : python , By : Joshua Johnson
Date : March 29 2020, 07:55 AM
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