New submission from Vykouk <[email protected]>:
The methods .get(), .pop(), .setdefault() evaluate defaults even though the key
already exists in the dictionary:
# -*- coding: utf-8 -*-
def deflt(x):
print('\nKey', x, 'is not in the dictionary')
return 'Default'
dicti = {1:'one', 2:'two', 3:'three'}
key = 2
print('\ndicti \t\t', dicti)
print('\t\t key =',key)
print('get \t\t', dicti.get(key, deflt(key)))
print('setdefault \t',dicti.setdefault(key, deflt(key)))
print('dicti \t\t', dicti)
print('pop \t\t', dicti.pop(key, deflt(key)))
print('dicti \t\t', dicti)
print('setdefault \t',dicti.setdefault(key, deflt(key)))
print('dicti \t\t', dicti)
'''
Printed outputs:
Python 3.7.1 (v3.7.1:260ec2c36a, Oct 20 2018, 14:57:15) [MSC v.1915 64 bit
(AMD64)]
Type "copyright", "credits" or "license" for more information.
IPython 7.1.1 -- An enhanced Interactive Python.
runfile('C:/Temp/Smazat/Dictionary_Defaults.py', wdir='C:/Temp/Smazat')
dicti {1: 'one', 2: 'two', 3: 'three'}
key = 2
Key 2 is not in the dictionary # ???
get two
Key 2 is not in the dictionary # ???
setdefault two
dicti {1: 'one', 2: 'two', 3: 'three'}
Key 2 is not in the dictionary # ???
pop two
dicti {1: 'one', 3: 'three'}
Key 2 is not in the dictionary
setdefault Default
dicti {1: 'one', 3: 'three', 2: 'Default'}
'''
----------
files: Dictionary_Defaults.py
messages: 343362
nosy: vykouk
priority: normal
severity: normal
status: open
title: Dictionary defaults .get(), .pop(), .setdefault()
type: behavior
versions: Python 3.7
Added file: https://bugs.python.org/file48352/Dictionary_Defaults.py
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Python tracker <[email protected]>
<https://bugs.python.org/issue37033>
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