Please is there any other way to write the line l = [x for x in l if x not in n] and write the code another way ? Please help me on python without using "break" or "continue", and no global variables in functions.
Please is there any other way to write the line l = [x for x in l if x not in n] and write the code another way ? Please help me on python without using "break" or "continue", and no global variables in functions.
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Please is there any other way to write the line l = [x for x in l if x not in n] and write the code another way ? Please help me on python without using "break" or "continue", and no global variables in functions.
![def remove_pairs(1):
(list of str)->list of str
Returns a copy of list 1 where all the pairs from 1 are removed AND
the elements of the new list shuffled
Precondition: elements of 1 are cards represented as strings described above
Testing:
Note that for the individual calls below, the function should
return the displayed list but not necessarily in the order given in the examples.
>>> remove_pairs(['94', '54',
['104', '24', '3♡',
>>> remove_pairs(['104'
['24', '50',
'44'
'54', '7♡', 'A^', '104', 'Q♡', '8♡'.
190', '100
'J♡', '10♡',
'J h', '3♡'])
'KO', 'A',
'A',
6',
'KO',
'J4', 'Q0']
'AO',
'24', 'Q*', 'KA', 'Qo', 'Jª', 'A♡',
'44',
'7♡',
'80',
194',
'24', '50',
'AO'j
'100'])
'6', '94'
no_pairs=[]
for i in ('2','3','4','5','6','7','8','9','10','J','Q','K','A'):
n = []
for j in range (len(1)):
if 1[j].find (i) != -1:
n.append (1[j])
if len(n) == 2 or len(n) == 3:
1.remove (n[0])
1.remove (n[1])
elif len(n) == 4:
1 = [x for x in 1 if x not in n]
no_pairs = 1
random. shuffle(no_pairs)
return no_pairs](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdb039aed-b356-40cc-ba3e-936586ecb4b2%2F30e3cc74-3c81-4990-a79b-783a5858e4d2%2Fckt5wd_processed.png&w=3840&q=75)
Transcribed Image Text:def remove_pairs(1):
(list of str)->list of str
Returns a copy of list 1 where all the pairs from 1 are removed AND
the elements of the new list shuffled
Precondition: elements of 1 are cards represented as strings described above
Testing:
Note that for the individual calls below, the function should
return the displayed list but not necessarily in the order given in the examples.
>>> remove_pairs(['94', '54',
['104', '24', '3♡',
>>> remove_pairs(['104'
['24', '50',
'44'
'54', '7♡', 'A^', '104', 'Q♡', '8♡'.
190', '100
'J♡', '10♡',
'J h', '3♡'])
'KO', 'A',
'A',
6',
'KO',
'J4', 'Q0']
'AO',
'24', 'Q*', 'KA', 'Qo', 'Jª', 'A♡',
'44',
'7♡',
'80',
194',
'24', '50',
'AO'j
'100'])
'6', '94'
no_pairs=[]
for i in ('2','3','4','5','6','7','8','9','10','J','Q','K','A'):
n = []
for j in range (len(1)):
if 1[j].find (i) != -1:
n.append (1[j])
if len(n) == 2 or len(n) == 3:
1.remove (n[0])
1.remove (n[1])
elif len(n) == 4:
1 = [x for x in 1 if x not in n]
no_pairs = 1
random. shuffle(no_pairs)
return no_pairs
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