Write the function mutatingRemoveRepeats (L) which takes a list L, and removes all the repeated integers from L so only the first instance of each of those repeated integers remains. For example, if L = [1, 5, 2, 2, 4, 4, 21. then after a call to (L), then L == [1, 5, 2, 4]. mutatingRemoveRepeats This is a mutating function, so it directly changes the provided list. You may assume that L contains only integers. As with most mutating functions, this function returns None. Also write the function nonmutatingRemoveRepeats (L), which works the same as the previous function, only this version is non-mutating. It does not modify L., and it returns a new list without any repeated values. Note: You may not simply call mutatingRemoveRepeats (L) in nonmutatingRemoveRepeats (L). Thus, instead of copying the list and then mutating that copy to get the answer, you need to build up a new list from scratch, like we did in the notes.

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
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4.2.16 removeRepeats (mutating and nonmutating)
EF D E F Q
from cmu_cs3_utils import testFunction
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3 def mutatingRemoveRepeats (L):
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return 42
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10 def testMutatingRemoveRepeats():
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L = [1, 5, 2, 2, 4, 4, 21
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def nonmutatingRemoveRepeats (L):
return 42
@testFunction
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mutatingRemoveRepeats (L)
assert (L == [1, 5, 2, 4])
L = [-2, 15, -2, 3]
mutatingRemoveRepeats (L)
assert (L= [-2, 15, 3])
L = [1, 1, 1]
mutatingRemoveRepeats (L)
assert (L [1])
L = [1, 9, 5]
mutatingRemoveRepeats (L)
assert (L == [1, 9, 5])
L = []
mutatingRemoveRepeats (L)
assert (L= [])
assert (mutatingRemoveRepeats (L) == None)
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33 @testFunction
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41-
def testNonmutatingRemoveRepeats():
assert (nonmutatingRemoveRepeats ([1, 5, 2, 2, 4, 4, 2]) == [1, 5, 2, 4])
assert (nonmutatingRemove Repeats([-2, 15, -2, 3]) == [-2, 15, 31)
assert(nonmutatingRemoveRepeats([1, 1, 1]) == [1])
assert (nonmutatingRemoveRepeats([1, 9, 5]) == [1, 9, 5])
assert (nonmutatingRemoveRepeats([]) == [])
#Verify that the function is nonmutating:
L = [1, 2, 2, 3, 4, 4]
nonmutatingRemoveRepeats (L)
assert (L = [1, 2, 2, 3, 4, 4])
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46 def main():
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testMutatingRemoveRepeats()
testNonmutatingRemoveRepeats()
main()
Stop
C Reset
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Write the function mutating RemoveRepeats (L) which takes a list L, and removes all
the repeated integers from L so only the first instance of each of those repeated integers
remains. For example, if L = [1, 5, 2, 2, 4, 4, 2], then after a call to
mutatingRemoveRepeats (L), then L == [1, 5, 2, 4].
This is a mutating function, so it directly changes the provided list. You may assume that L
contains only integers. As with most mutating functions, this function returns None.
Also write the function nonmutatingRemoveRepeats (L), which works the same as the
previous function, only this version is non-mutating. It does not modify L, and it returns a
new list without any repeated values.
Note: You may not simply call mutatingRemoveRepeats (L) in
nonmutatingRemoveRepeats (L). Thus, instead of copying the list L and then mutating
that copy to get the answer, you need to build up a new list from scratch, like we did in the
notes.
Transcribed Image Text:4.2.16 removeRepeats (mutating and nonmutating) EF D E F Q from cmu_cs3_utils import testFunction 1 2 3 def mutatingRemoveRepeats (L): 4 return 42 5 6 7 8 9 10 def testMutatingRemoveRepeats(): 11 L = [1, 5, 2, 2, 4, 4, 21 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 def nonmutatingRemoveRepeats (L): return 42 @testFunction 42 43 44 mutatingRemoveRepeats (L) assert (L == [1, 5, 2, 4]) L = [-2, 15, -2, 3] mutatingRemoveRepeats (L) assert (L= [-2, 15, 3]) L = [1, 1, 1] mutatingRemoveRepeats (L) assert (L [1]) L = [1, 9, 5] mutatingRemoveRepeats (L) assert (L == [1, 9, 5]) L = [] mutatingRemoveRepeats (L) assert (L= []) assert (mutatingRemoveRepeats (L) == None) 31 32 33 @testFunction 34 35 36 37 38 39 40 41- def testNonmutatingRemoveRepeats(): assert (nonmutatingRemoveRepeats ([1, 5, 2, 2, 4, 4, 2]) == [1, 5, 2, 4]) assert (nonmutatingRemove Repeats([-2, 15, -2, 3]) == [-2, 15, 31) assert(nonmutatingRemoveRepeats([1, 1, 1]) == [1]) assert (nonmutatingRemoveRepeats([1, 9, 5]) == [1, 9, 5]) assert (nonmutatingRemoveRepeats([]) == []) #Verify that the function is nonmutating: L = [1, 2, 2, 3, 4, 4] nonmutatingRemoveRepeats (L) assert (L = [1, 2, 2, 3, 4, 4]) 45 46 def main(): 47 48 49 50 testMutatingRemoveRepeats() testNonmutatingRemoveRepeats() main() Stop C Reset Check Code History Run ● Saved Write the function mutating RemoveRepeats (L) which takes a list L, and removes all the repeated integers from L so only the first instance of each of those repeated integers remains. For example, if L = [1, 5, 2, 2, 4, 4, 2], then after a call to mutatingRemoveRepeats (L), then L == [1, 5, 2, 4]. This is a mutating function, so it directly changes the provided list. You may assume that L contains only integers. As with most mutating functions, this function returns None. Also write the function nonmutatingRemoveRepeats (L), which works the same as the previous function, only this version is non-mutating. It does not modify L, and it returns a new list without any repeated values. Note: You may not simply call mutatingRemoveRepeats (L) in nonmutatingRemoveRepeats (L). Thus, instead of copying the list L and then mutating that copy to get the answer, you need to build up a new list from scratch, like we did in the notes.
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