Google Python Class | List Exercise –
Given a list of numbers, return a list where
all adjacent == elements have been reduced to a single element,
so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or
modify the passed in list.
My solution using a new list is –
def remove_adjacent(nums):
a = []
for item in nums:
if len(a):
if a[-1] != item:
a.append(item)
else: a.append(item)
return a
The question even suggests that it could be done by modifying the passed in list. However, the python documentation warned against modifying elements while iterating a list using the for loop.
I am wondering what else can I try apart from iterating over the list, to get this done. I am not looking for the solution, but maybe a hint that can take me into a right direction.
UPDATE
-updated the above code with suggested improvements.
-tried the following with a while loop using suggested hints –
def remove_adjacent(nums):
i = 1
while i < len(nums):
if nums[i] == nums[i-1]:
nums.pop(i)
i -= 1
i += 1
return nums
Answers:
Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.
Method 1
Here’s the traditional way, deleting adjacent duplicates in situ, while traversing the list backwards:
Python 1.5.2 (#0, Apr 13 1999, 10:51:12) [MSC 32 bit (Intel)] on win32 Copyright 1991-1995 Stichting Mathematisch Centrum, Amsterdam >>> def dedupe_adjacent(alist): ... for i in xrange(len(alist) - 1, 0, -1): ... if alist[i] == alist[i-1]: ... del alist[i] ... >>> data = [1,2,2,3,2,2,4]; dedupe_adjacent(data); print data [1, 2, 3, 2, 4] >>> data = []; dedupe_adjacent(data); print data [] >>> data = [2]; dedupe_adjacent(data); print data [2] >>> data = [2,2]; dedupe_adjacent(data); print data [2] >>> data = [2,3]; dedupe_adjacent(data); print data [2, 3] >>> data = [2,2,2,2,2]; dedupe_adjacent(data); print data [2] >>>
Update: If you want a generator but (don’t have itertools.groupby or (you can type faster than you can read its docs and understand its default behaviour)), here’s a six-liner that does the job:
Python 2.3.5 (#62, Feb 8 2005, 16:23:02) [MSC v.1200 32 bit (Intel)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> def dedupe_adjacent(iterable): ... prev = object() ... for item in iterable: ... if item != prev: ... prev = item ... yield item ... >>> data = [1,2,2,3,2,2,4]; print list(dedupe_adjacent(data)) [1, 2, 3, 2, 4] >>>
Update 2: Concerning the baroque itertools.groupby() and the minimalist object() …
To get the dedupe_adjacent effect out of itertools.groupby(), you need to wrap a list comprehension around it to throw away the unwanted groupers:
>>> [k for k, g in itertools.groupby([1,2,2,3,2,2,4])] [1, 2, 3, 2, 4] >>>
… or muck about with itertools.imap and/or operators.itemgetter, as seen in another answer.
Expected behaviour with object instances is that none of them compares equal to any other instance of any class, including object itself. Consequently they are extremely useful as sentinels.
>>> object() == object() False
It’s worth noting that the Python reference code for itertools.groupby uses object() as a sentinel:
self.tgtkey = self.currkey = self.currvalue = object()
and that code does the right thing when you run it:
>>> data = [object(), object()] >>> data [<object object at 0x00BBF098>, <object object at 0x00BBF050>] >>> [k for k, g in groupby(data)] [<object object at 0x00BBF098>, <object object at 0x00BBF050>]
Update 3: Remarks on forward-index in-situ operation
The OP’s revised code:
def remove_adjacent(nums):
i = 1
while i < len(nums):
if nums[i] == nums[i-1]:
nums.pop(i)
i -= 1
i += 1
return nums
is better written as:
def remove_adjacent(seq): # works on any sequence, not just on numbers
i = 1
n = len(seq)
while i < n: # avoid calling len(seq) each time around
if seq[i] == seq[i-1]:
del seq[i]
# value returned by seq.pop(i) is ignored; slower than del seq[i]
n -= 1
else:
i += 1
#### return seq #### don't do this
# function acts in situ; should follow convention and return None
Method 2
Use a generator to iterate over the elements of the list, and yield a new one only when it has changed.
itertools.groupby does exactly this.
You can modify the passed-in list if you iterate over a copy:
for elt in theList[ : ]:
...
Method 3
Just to show one more way here is another single liner version without indexes:
def remove_adjacent(nums):
return [a for a,b in zip(nums, nums[1:]+[not nums[-1]]) if a != b]
The not part puts the last value to result as only a ends up to result.
Method 4
As usual, I am just here to advertise the impressive recipes in the Python itertools documentation.
What you are looking for is the function unique_justseen:
from itertools import imap, groupby
from operator import itemgetter
def unique_justseen(iterable, key=None):
"List unique elements, preserving order. Remember only the element just seen."
# unique_justseen('AAAABBBCCDAABBB') --> A B C D A B
# unique_justseen('ABBCcAD', str.lower) --> A B C A D
return imap(next, imap(itemgetter(1), groupby(iterable, key)))
list(unique_justseen([1,2,2,3])) # [1, 2, 3]
Method 5
Well, katrielalex is right about itertools, but the OP seems to be rather more interested (or should be!) in learning to manipulate the basics of the built-in data structures. As for manipulating a list in place, it does need thought, but my recommendation would be to read through this section of the documentation and try a few list methods (hint: list.pop(), list.remove(), and learn everything about slices.)
The posted code could be simplified, by the way (you should however add handling of error conditions):
def remove_adjacent(nums):
a = nums[:1]
for item in nums[1:]:
if item != a[-1]:
a.append(item)
return a
Method 6
Extremely elegant solution from Google (source is here: https://developers.google.com/edu/python/exercises/basic):
def remove_adjacent(nums):
result = []
for num in nums:
if len(result) == 0 or num != result[-1]:
result.append(num)
return result
Method 7
You can use list comprehension. For example something like this should do the job:
def remove_adjacent(L): return [elem for i, elem in enumerate(L) if i == 0 or L[i-1] != elem]
or:
def remove_adjacent(L): return [L[i] for i in xrange(len(L)) if i == 0 or L[i-1] != L[i]]
Method 8
Try this:
def remove_adjacent(nums):
result = []
if len(nums) > 0:
result = [nums[0]]
for i in range(len(nums)-1):
if nums[i] != nums[i+1]:
result.append(nums[i+1])
return result
Method 9
itertools.groupby is superior, but there is also
reduce(lambda x, y: x + [y] if x[-1] != y else x, seq[1:], seq[0:1])
e.g.
>>> seq = [[1,1], [2,2], [3,3], [3,3], [2,2], [2,2], [1,1]] >>> print reduce(lambda x, y: x + [y] if x[-1] != y else x, seq[1:], seq[0:1]) [[1, 1], [2, 2], [3, 3], [2, 2], [1, 1]]
When coming from functional languages where this sort of thing is done with a fold, then using reduce often feels natural.
Method 10
You can modify a list you’re iterating over if you use indices explicitly:
def remove_adjacent(l):
if len(l)<2:
return l
prev,i = l[0],1
while i < len(l):
if l[i] == prev:
del l[i]
else:
prev = l[i]
i += 1
It doesn’t work with iterators because iterators don’t “know” how to modify the index when you remove arbitrary elements, so it’s easier to just forbid it. Some languages have iterators with functions to remove the “current item”.
Method 11
@katrielalex’s solution is more pythonic, but if you did need to modify the list in-place without making a copy, you could use a while loop and break when you catch an IndexError.
e.g.
nums = [1,1,1,2,2,3,3,3,5,5,1,1,1]
def remove_adjacent(nums):
"""Removes adjacent items by modifying "nums" in-place. Returns None!"""
i = 0
while True:
try:
if nums[i] == nums[i+1]:
# Letting you figure this part out,
# as it's a homework question
except IndexError:
break
print nums
remove_adjacent(nums)
print nums
Edit: pastebin of one way to do it here, in case you get stuck and want to know..
Method 12
def remove_adjacent(nums):
newList=[]
for num in nums:
if num not in newList:
newList.append(num)
newList.sort()
return newList
Method 13
Another approach. Comments welcome.
def remove_adjacent(nums):
'''modifies the list passed in'''
l, r = 0, 1
while r < len(nums):
if nums[l] == nums[r]:
r += 1
else:
l += 1
nums[l] = nums[r]
r += 1
del nums[l+1:]
Method 14
Seeing the code written by Google is a humbling lol. This is what I came up with:
def remove_adjacent(nums):
rmvelement = []
checkedIndex = []
for num in nums:
if nums.index(num) not in checkedIndex:
index = nums.index(num)
checkedIndex.append(index)
skip = False
else:
skip = True
if skip == False:
for x in nums[index+1:]:
if x == num:
rmvelement.append(x)
else:
break
[nums.remove(_) for _ in rmvelement]
return nums
Method 15
This should work for a transparent (albeit roundabout) solution:
def remove_adjacent(nums):
numstail = [i for i in range(0,len(nums))]
nums = nums + numstail
for i in nums:
if nums[i] == nums[i-1]:
del nums[i]
return nums[:-len(numstail)]
The logic is as follows:
- Create a tail-list equal to the length of the original list of numbers and append this to the end of the original list.
- Run a ‘for-loop’ that checks if a given element of nums is the same as the previous element. If so, delete it.
- Return the new nums list, with the necessary deletions, up to
len(numtails)index positions from the end of the list.
(numstail is defined to avoid indices being out of range for any length list)
Method 16
def removeDupAdj2(a):
b=[]
for i in reversed(range(1,len(a))):
if(a[i-1] == a[i]):
del(a[i])
#print(a)
return a
a = [int(i) for i in '1 2 3 3 4 4 3 5 4 4 6 6 6 7 8 8 8 9 1 1 0 0'.split(' ')]
a
res = removeDupAdj2(a)
res
Method 17
Since you are in a Python class, I’m going to guess that you are new to the language. Thus, for you and any other beginners out there, I wrote a simple version of the code to help others get through the logic.
original= [1, 2, 2, 3]
newlist=[]
for item in original:
if item in newlist:
print "You don't need to add "+str(item)+" again."
else:
newlist.append(item)
print "Added "+str(item)
print newlist
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0