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Why inserting keys in order into a python dict is faster than doint it unordered


Why inserting keys in order into a python dict is faster than doint it unordered

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Why inserting keys in order into a python dict is faster than doint it unordered
Tag : python , By : user103892
Date : November 29 2020, 01:01 AM


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Order of keys in a different Python dict()


Tag : python , By : Sandeep Arneja
Date : March 29 2020, 07:55 AM
I wish this helpful for you , You cannot rely on the key order at all:
>>> {1: None, 9: None}
{1: None, 9: None}
>>> {9: None, 1: None}
{9: None, 1: None}
>>> {1: None, 2: None}
{1: None, 2: None}
>>> {2: None, 1: None}
{1: None, 2: None}

Python is 'key in dict' different/faster than 'key in dict.keys()'


Tag : python , By : user121350
Date : March 29 2020, 07:55 AM
will help you Short answer:
In python 2: your assumption is correct: dict.keys() slows down. In python 3: your assumption is not correct: in dict.keys() performs like like in dict
#! /usr/bin/python2.7

import datetime as dt
import random

dict_size = 1000000
num_iterations = 100

d = {i: i for i in xrange(dict_size)}

def f():
    k = random.randint(0,dict_size-1)
    return (k in d)

def g():
    k = random.randint(0,dict_size-1)
    return (k in d.keys())

def test(func):
    t = dt.datetime.utcnow()
    for i in xrange(num_iterations):
        func()
    print "%s --> %1.6f s" % (func, (dt.datetime.utcnow()-t).total_seconds())

test(f)
test(g)
<function f at 0x7ff2e0126d70> --> 0.000598 s
<function g at 0x7ff2e0126de8> --> 5.191553 s
<function f at 0x7f94cb5e6d70> --> 3.614162 s
<function g at 0x7f94cb5e6de8> --> 7.007922 s

Iterate over a column containing keys from a dict. Return matched keys from second dict keeping order of keys from first


Tag : python , By : Chris Tattum
Date : March 29 2020, 07:55 AM
hop of those help? As I see, you imported OrderedDict, but didn't use it. You should build OrderedDict to save keys order:
dict_a = OrderedDict((rows[1],rows[2]) for rows in reader)
dict_b = dict((rows[3],rows[4]) for rows in reader)

for key, value in dict_a.iteritems():
    if dict_b[key] == value:
        print value

Why is the order of python dict keys not consistent?


Tag : python , By : Arun Thakkar
Date : March 29 2020, 07:55 AM
wish helps you This behavior is detailed in object.__hash__()'s specification; it's to prevent certain types of malicious input from breaking applications:

Need help iterating through Python dict keys/values and INSERTing into SQL DB


Tag : python , By : user150694
Date : March 29 2020, 07:55 AM
To fix the issue you can do Consider unpacking your collection of dictionaries into key/value tuples and then parameterize the values tuple in the loop. Assuming the below data structure (list of dictionaries):
scale_data_json["operations"] = [{'BMI': 0, 'BodyFat': 10, 
                                  'Entrytimestamp': '2018-01-21T19:37:47.821Z', 
                                  'MuscleMass': 50, 'OperationType': 'create',
                                  'ServerTimestamp':'2018-01-21T19:37:47.821Z', 
                                  'Source':'bluetooth scale', 
                                  'Water':37, 'Weight':21},
                                 {'BMI': 0, 'BodyFat': 10, 
                                  'Entrytimestamp': '2018-01-21T19:37:47.821Z', 
                                  'MuscleMass': 50, 'OperationType': 'create',
                                  'ServerTimestamp':'2018-01-21T19:37:47.821Z', 
                                  'Source':'bluetooth scale', 
                                  'Water':37, 'Weight':21},
                                ...]
# PREPARED STATEMENT
sql = """INSERT INTO BodyComposition (BMI, BodyFat, Entrytimestamp, 
                                      MuscleMass, OperationType, ServerTimestamp, 
                                      Source, Water, Weight) 
         VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?)
      """

# LOOP, UNPACK, BIND PARAMS
for entry in scale_data_json["operations"]:
    keys, values = zip(*entry.items())
    cursor.execute(sql, values)
    cnxn.commit()
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