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问题:
I'm supposed to create a function, which input is a list and two numbers, the function reverses the sublist which its place is indicated by the two numbers.
for example this is what it's supposed to do:
>>> lst = [1, 2, 3, 4, 5]
>>> reverse_sublist (lst,0,4)
>>> lst [4, 3, 2, 1, 5]
I created a function and it works, but I'm not sure is it's in place.
This is my code:
def reverse_sublist(lst,start,end):
sublist=lst[start:end]
sublist.reverse()
lst[start:end]=sublist
print(lst)
回答1:
def reverse_sublist(lst,start,end):
lst[start:end] = lst[start:end][::-1]
return lst
回答2:
Partial reverse with no temporary list (replace range
with xrange
if you use Python 2):
def partial_reverse(list_, from_, to):
for i in range(0, int((to - from_)/2)):
(list_[from_+i], list_[to-i]) = (list_[to-i], list_[from_+i])
list_ = [1, 2, 3, 4, 5, 6, 7, 8]
partial_reverse(list_, 3, 7)
print(list_)
回答3:
Easiest way to reverse a list in a partial or complete manner.
listVar = ['a','b','c','d']
def listReverse(list,start,end):
while(start<end):
temp = list[start]
list[start] = list[end] #Swaping
list[end]=temp
start+=1
end-=1
print(list)
listReverse(listVar,1,3)
Output : - ['a', 'd', 'c', 'b']
回答4:
... I'm not sure is it's in place.
...
lst[start:end]=sublist
Yes, it's in place. lst
is never rebound, only its object mutated.
回答5:
Try some crazy slicing
, see Explain Python's slice notation and http://docs.python.org/2.3/whatsnew/section-slices.html
x = [1,2,3,4,5,6,7,8]
def sublist_reverse(start_rev, end_rev, lst):
return lst[:end_rev-1:start_rev-1]+lst[:[end_rev]
print sublist_reverse(0,4,x)
[out]:
[8, 7, 6, 5, 4, 3, 2, 1]
回答6:
Not sure if you have a similar problem as mine, but i needed to reverse a list in place.
The only piece I was missing was [:]
exStr = "String"
def change(var):
var[:] = var[::-1] # This line here
print(exStr) #"String"
change(exStr)
print(exStr) #"gnirtS"
回答7:
Just use a slice:
>>> lst = [1, 2, 3, 4, 5]
>>> lst[0:len(lst[3::-1])]=lst[3::-1]
>>> lst
[4, 3, 2, 1, 5]
Or, perhaps easier to understand:
>>> lst = [1, 2, 3, 4, 5]
>>> sl=lst[3::-1]
>>> lst[0:len(sl)]=sl
>>> lst
[4, 3, 2, 1, 5]
回答8:
lst[::-1]
is the idiomatic way to reverse a list in Python, The following show how and that it was in-place:
>>> lst = [1, 2, 3, 4, 5]
>>> id(lst)
12229328
>>> lst[:] = lst[::-1]
>>> lst
[5, 4, 3, 2, 1]
>>> id(lst)
12229328
回答9:
I have two ways for in-place reversal, the simple way is to loop through the list half-way, swapping the elements with the respective mirror-elements. By mirror-element I mean (first, last), (2nd, 2nd-last), (3rd, 3rd-last), etc.
def reverse_list(A):
for i in range(len(A) // 2): # half-way
A[i], A[len(A) - i - 1] = A[len(A) - i - 1], A[i] #swap
return A
The other way is similar to the above but using recursion as opposed to a "loop":
def reverse_list(A):
def rev(A, start, stop):
A[start], A[stop] = A[stop], A[start] # swap
if stop - start > 1: # until halfway
rev(A, start + 1, stop - 1)
return A
return rev(A, 0, len(A) - 1)
回答10:
I've conducted a tiny experiment and it seems that any assignment to list slice causes memory allocation:
import resource
resource.setrlimit(resource.RLIMIT_AS, (64 * 1024, 64 * 1024))
try:
# Python 2
zrange = xrange
arr_size = 3 * 1024
except NameError:
# Python 3
zrange = range
arr_size = 4 * 1024
arr = list(zrange(arr_size))
# We could allocate additional 100 integers, so there should be enough free memory
# to allocate a couple of variables for indexes in the statement below
# additional_memory = list(zrange(100))
# MemoryError is raised here
arr[:] = zrange(arr_size)
So you have to use for loop to reverse a sublist in place.
PS: If you want to repeat this test, you should ensure that setrlimit RLIMIT_AS works fine on your platform. Also arr_size may vary for different python implementations.
回答11:
lst = [1,2,3,4,5,6,7,8]
Suppose you have to reverse 2nd position to 4th position in place.
lst[2:5] = lst[2:5][::-1]
Output:
[1,2,5,4,3,6,7,8]