I have an ordered list e.g. x = [0, 100, 200, 1000], and given any
positive integer y, I want to determine its appropriate position in
the list (i.e the point at which I would have to insert it in order to
keep the list sorted. I can clearly do this with a series of if
statements:
if y<x[1]:
n = 0
elif y < x[2]:
n = 1
elif y < x[3]:
n = 2
else:
n = 3
Or with a generator comprehension
n = sum ( y>x[i] for i in range(len(x)) ) - 1
But there has to be a cleaner way, as the first approach is unwieldy
and does not adapt to changing list lengths, and the second is not
obvious to a casual reader of the code.
My list will typically have 2 to 5 items, so speed is not a huge
issue. I'd appreciate your guidance.
Sincerely
Thomas Philips 19 2042
You might look at the bisect module (part of the standard
distribution).
On Mar 16, 12:59 pm, tkp...@hotmail.com wrote:
I have an ordered list e.g. x = [0, 100, 200, 1000], and given any
positive integer y, I want to determine its appropriate position in
the list (i.e the point at which I would have to insert it in order to
keep the list sorted. I can clearly do this with a series of if
statements:
if y<x[1]:
n = 0
elif y < x[2]:
n = 1
elif y < x[3]:
n = 2
else:
n = 3
Or with a generator comprehension
n = sum ( y>x[i] for i in range(len(x)) ) - 1
But there has to be a cleaner way, as the first approach is unwieldy
and does not adapt to changing list lengths, and the second is not
obvious to a casual reader of the code.
My list will typically have 2 to 5 items, so speed is not a huge
issue. I'd appreciate your guidance.
Sincerely
Thomas Philips
One way to do this would be to use the cmp built-in and loop over the
items in the list. Maybe something like this:
x = [0, 100, 200, 1000]
numLst = len(x)
count = 0
for i in range(numLst):
resultOfCmp = cmp(newNum, x[count])
if resultOfCmp == -1:
print i
x.insert(count, newNum)
break
count += 1
# Where newNum is the one to be inserted.
It's a hack, but it might get the ol' creative juices flowing.
Mike
On Mar 16, 12:59 pm, tkp...@hotmail.com wrote:
I have an ordered list e.g. x = [0, 100, 200, 1000], and given any
positive integer y, I want to determine its appropriate position in
the list (i.e the point at which I would have to insert it in order to
keep the list sorted. I can clearly do this with a series of if
statements:
if y<x[1]:
n = 0
elif y < x[2]:
n = 1
elif y < x[3]:
n = 2
else:
n = 3
Or with a generator comprehension
n = sum ( y>x[i] for i in range(len(x)) ) - 1
But there has to be a cleaner way, as the first approach is unwieldy
and does not adapt to changing list lengths, and the second is not
obvious to a casual reader of the code.
My list will typically have 2 to 5 items, so speed is not a huge
issue. I'd appreciate your guidance.
Sincerely
Thomas Philips
List "will typically have 2 to 5 items"? Keep it simple!
x.append(y)
x.sort()
-- Paul
On Mar 16, 2:32 pm, "Paul McGuire" <p...@austin.rr.comwrote:
On Mar 16, 12:59 pm, tkp...@hotmail.com wrote:
I have an ordered list e.g. x = [0, 100, 200, 1000], and given any
positive integer y, I want to determine its appropriate position in
the list (i.e the point at which I would have to insert it in order to
keep the list sorted. I can clearly do this with a series of if
statements:
if y<x[1]:
n = 0
elif y < x[2]:
n = 1
elif y < x[3]:
n = 2
else:
n = 3
Or with a generator comprehension
n = sum ( y>x[i] for i in range(len(x)) ) - 1
But there has to be a cleaner way, as the first approach is unwieldy
and does not adapt to changing list lengths, and the second is not
obvious to a casual reader of the code.
My list will typically have 2 to 5 items, so speed is not a huge
issue. I'd appreciate your guidance.
Sincerely
Thomas Philips
List "will typically have 2 to 5 items"? Keep it simple!
x.append(y)
x.sort()
-- Paul
I thought doing an append and sort was a good idea too, but the
question entailed knowing the insertion point, so I skipped it.
Thanks!
How about:
-----------
x = [0, 100, 200, 1000]
y = -1
inserted = False
for i in range(len(x)):
if(y <= x[i]):
x.insert(i, y)
inserted = True
break
if(not inserted): x.append(y)
print x
------------
On Mar 16, 11:20 am, "Matimus" <mccre...@gmail.comwrote:
You might look at the bisect module (part of the standard
distribution).
Here is an example:
>>from bisect import insort x = [0,100,200,1000] insort(x,10) x
[0, 10, 100, 200, 1000]
Or like this:
x = [0, 100, 200, 1000]
y = 435
for n, i in enumerate(x):
if y < i:
n = n - 1
break
x.insert(n + 1, y)
If you decide to stick with
n = sum ( y>x[i] for i in range(len(x)) ) - 1
Replace it with:
n = sum(y i for i in x) - 1
Tobias K. tk****@hotmail.com writes:
Or with a generator comprehension
n = sum ( y>x[i] for i in range(len(x)) ) - 1
But there has to be a cleaner way, as the first approach is unwieldy
and does not adapt to changing list lengths, and the second is not
obvious to a casual reader of the code.
How about :
n = len([y t for t in x])
On Fri, 16 Mar 2007 18:17:08 -0800, Paul Rubin wrote: tk****@hotmail.com writes:
>Or with a generator comprehension n = sum ( y>x[i] for i in range(len(x)) ) - 1
But there has to be a cleaner way, as the first approach is unwieldy and does not adapt to changing list lengths, and the second is not obvious to a casual reader of the code.
How about:
n = len([y t for t in x])
(1) It's wrong. That always returns the length of the list. Perhaps you
meant something like this?
len(["anything will do" for t in x if y t])
or even
len(filter(lambda t, y=y: y>t, x))
(2) It's barely more comprehensible than the alternative that the Original
Poster rejected for being insufficiently clear.
Since (almost) everyone insists on ignoring the bisect module and
re-inventing the wheel, here's my wheel:
def find(alist, n):
"""Return the position where n should be inserted in a sorted list."""
if alist != sorted(alist):
raise ValueError('list must be sorted')
where = None
for i in range(len(alist)):
if where is not None:
break
alist.insert(i, n)
if alist == sorted(alist):
where = i
del alist[i]
if where is None:
where = len(alist)
return where
Here's another dodgy implementation:
def find(alist, n):
return sorted(alist + [n]).index(n)
It's especially good for large lists. Not!
--
Steven.
Steven D'Aprano <st***@REMOVE.THIS.cybersource.com.auwrites:
(1) It's wrong. That always returns the length of the list. Perhaps you
meant something like this?
len(["anything will do" for t in x if y t])
Yeah, that's what I meant.
Steven D'Aprano <st***@REMOVE.THIS.cybersource.com.auwrites:
or even
len(filter(lambda t, y=y: y>t, x))
How about
min(i for i,t in enumerate(x) if t >= y)
or
max(i for i,t in enumerate(x) if t <= y)
Those are actually pretty direct.
"7stud" schrieb
How about:
-----------
x = [0, 100, 200, 1000]
y = -1
inserted = False
for i in range(len(x)):
if(y <= x[i]):
x.insert(i, y)
inserted = True
break
if(not inserted): x.append(y)
print x
------------
You can get rid of the sentinel "inserted" using the
else clause of the for loop:
for i in range(len(x)):
if (y <= x[i]):
x.insert(i, y)
break
else: x.append(y)
Python is cool :-)
IMHO. HTH.
Martin
On Mar 17, 5:42 pm, Paul Rubin <http://phr...@NOSPAM.invalidwrote:
Steven D'Aprano <s...@REMOVE.THIS.cybersource.com.auwrites:
or even
len(filter(lambda t, y=y: y>t, x))
How about
min(i for i,t in enumerate(x) if t >= y)
or
max(i for i,t in enumerate(x) if t <= y)
Those are actually pretty direct.
I'd hate to see "indirect". Worse, the min-using gizmoid crashes when
y x[-1] -- all your ifs are belong to False.
>>x
[0, 100, 200, 1000]
>>tests = [0, 1, 100, 150, 1000, 2000] [(y, max(i for i,t in enumerate(x) if t <= y)) for y in tests]
[(0, 0), (1, 0), (100, 1), (150, 1), (1000, 3), (2000, 3)]
Looks OK, iff one is happy with the OP's strange usage of "insert
point".
>>xc = x[:] xc.insert(1, 150) xc
[0, 150, 100, 200, 1000]
Whoops.
Try this for size:
>>[(y, sum(t <= y for t in x)) for y in tests]
[(0, 1), (1, 1), (100, 2), (150, 2), (1000, 4), (2000, 4)]
Cheers,
John
Paul Rubin wrote:
Steven D'Aprano <st***@REMOVE.THIS.cybersource.com.auwrites:
>or even
len(filter(lambda t, y=y: y>t, x))
How about
min(i for i,t in enumerate(x) if t >= y)
or
max(i for i,t in enumerate(x) if t <= y)
Those are actually pretty direct.
How about a solution (like the bisect one suggested almost as soon as
this thread started) that doesn't iterate over the whole list.
Having said which, for the promoted use cases I agree that the append()
and sort() paradigm wins hands down until it starts to make a real and
observable difference to the run time.
regards
Steve
--
Steve Holden +44 150 684 7255 +1 800 494 3119
Holden Web LLC/Ltd http://www.holdenweb.com
Skype: holdenweb http://del.icio.us/steve.holden
Recent Ramblings http://holdenweb.blogspot.com
On Mar 17, 9:46 pm, Steve Holden <s...@holdenweb.comwrote:
Paul Rubin wrote:
Steven D'Aprano <s...@REMOVE.THIS.cybersource.com.auwrites:
or even
len(filter(lambda t, y=y: y>t, x))
How about
min(i for i,t in enumerate(x) if t >= y)
or
max(i for i,t in enumerate(x) if t <= y)
Those are actually pretty direct.
How about a solution (like the bisect one suggested almost as soon as
this thread started) that doesn't iterate over the whole list.
Having said which, for the promoted use cases I agree that the append()
and sort() paradigm wins hands down until it starts to make a real and
observable difference to the run time.
Unfortunately for sort/append, the OP wants to find the insertion
point -- he hasn't mentioned actually doing the insertion.
On Mar 17, 4:12 am, "Martin Blume" <mbl...@socha.netwrote:
"7stud" schrieb
How about:
-----------
x = [0, 100, 200, 1000]
y = -1
inserted = False
for i in range(len(x)):
if(y <= x[i]):
x.insert(i, y)
inserted = True
break
if(not inserted): x.append(y)
print x
------------
You can get rid of the sentinel "inserted" using the
else clause of the for loop:
for i in range(len(x)):
if (y <= x[i]):
x.insert(i, y)
break
else: x.append(y)
Python is cool :-)
IMHO. HTH.
Martin
for-else? Neat.
Steve Holden <st***@holdenweb.comwrites:
max(i for i,t in enumerate(x) if t <= y)
Those are actually pretty direct.
How about a solution (like the bisect one suggested almost as soon as
this thread started) that doesn't iterate over the whole list.
Here's a Haskell-inspired one:
len(list(itertools.takewhile(lambda t: y t, x)))
It stops iterating when it hits an element >= y. I originally wanted
to write the above as:
len(itertools.takewhile(y.__gt__, x))
but it looks like regular numbers only support __cmp__ and not rich
comparison, and also you can't take the length of an iterator. In
Haskell this type of thing is very natural:
length(takeWhile (y >) x)
On Mar 18, 2:23 am, Paul Rubin <http://phr...@NOSPAM.invalidwrote:
Steve Holden <s...@holdenweb.comwrites:
max(i for i,t in enumerate(x) if t <= y)
Those are actually pretty direct.
How about a solution (like the bisect one suggested almost as soon as
this thread started) that doesn't iterate over the whole list.
Here's a Haskell-inspired one:
len(list(itertools.takewhile(lambda t: y t, x)))
Can you explain how list() works in that statement. I looked up
takewhile() and it returns an iterator that will automatically stop at
the insertion point? So does list() do an internal comprehension with
the iterator?
7stud <bb**********@yahoo.comwrote:
On Mar 18, 2:23 am, Paul Rubin <http://phr...@NOSPAM.invalidwrote:
Steve Holden <s...@holdenweb.comwrites:
max(i for i,t in enumerate(x) if t <= y)
Those are actually pretty direct.
How about a solution (like the bisect one suggested almost as soon as
this thread started) that doesn't iterate over the whole list.
Here's a Haskell-inspired one:
len(list(itertools.takewhile(lambda t: y t, x)))
Can you explain how list() works in that statement. I looked up
takewhile() and it returns an iterator that will automatically stop at
the insertion point? So does list() do an internal comprehension with
the iterator?
Any call to list(iterator) works roughly as follows:
def likelist(iterator):
result = []
while True:
try: result.append(iterator.next())
except StopIteration: return result
Alex This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics
by: Daniel Fortin |
last post by:
I was wondering if anyone had any suggestions for this problem:
I would like to use a map to store separate lists. Each list can have
a variable number of elements. The problem is that I would...
|
by: John N. |
last post by:
Hi All,
Here I have a linked list each containing a char and is double linked.
Then I have a pointer to an item in that list which is the current
insertion point.
In this funtion, the user...
|
by: chai |
last post by:
I am trying out a program to insert an element to a sorted list(singly
linked list)without using the temporary variable.Is there a solution
for this problem?
|
by: Sam Marrocco |
last post by:
I've created a collection class that inherits the
NameObjectCollectionBase. I'd like to add a few functions, such as being
able to insert a new item at an indexed or keyed location in the...
|
by: Joel Hedlund |
last post by:
Hi all!
I use python for writing terminal applications and I have been bothered
by how hard it seems to be to determine the terminal size. What is the
best way of doing this?
At the end I've...
|
by: Julia |
last post by:
I am trying to sort a linked list using insertion sort. I have seen a
lot of ways to get around this problem but no time-efficient and
space-efficient solution. This is what I have so far:
...
|
by: barcaroller |
last post by:
If I insert/remove an element in a set, will an iterator to this set
automatically become invalid? Does the position of the iterator before the
insertion/removal matter?
How are vectors and...
|
by: Jeff |
last post by:
As I'm learning PHP, I'm making a fair number of mistakes in syntax.
In perl, you can turn on reading these errors from the browser by
adding this:
use CGI::Carp 'fatalsToBrowser';
Is there...
|
by: python_newbie |
last post by:
I don't know this list is the right place for newbie questions. I try
to implement insertion sort in pyhton. At first code there is no
problem. But the second one ( i code it in the same pattern i...
|
by: ryjfgjl |
last post by:
If we have dozens or hundreds of excel to import into the database, if we use the excel import function provided by database editors such as navicat, it will be extremely tedious and time-consuming...
|
by: ryjfgjl |
last post by:
In our work, we often receive Excel tables with data in the same format. If we want to analyze these data, it can be difficult to analyze them because the data is spread across multiple Excel files...
|
by: emmanuelkatto |
last post by:
Hi All, I am Emmanuel katto from Uganda. I want to ask what challenges you've faced while migrating a website to cloud.
Please let me know.
Thanks!
Emmanuel
|
by: nemocccc |
last post by:
hello, everyone, I want to develop a software for my android phone for daily needs, any suggestions?
|
by: marktang |
last post by:
ONU (Optical Network Unit) is one of the key components for providing high-speed Internet services. Its primary function is to act as an endpoint device located at the user's premises. However,...
|
by: Hystou |
last post by:
Most computers default to English, but sometimes we require a different language, especially when relocating. Forgot to request a specific language before your computer shipped? No problem! You can...
|
by: jinu1996 |
last post by:
In today's digital age, having a compelling online presence is paramount for businesses aiming to thrive in a competitive landscape. At the heart of this digital strategy lies an intricately woven...
|
by: Hystou |
last post by:
Overview:
Windows 11 and 10 have less user interface control over operating system update behaviour than previous versions of Windows. In Windows 11 and 10, there is no way to turn off the Windows...
|
by: tracyyun |
last post by:
Dear forum friends,
With the development of smart home technology, a variety of wireless communication protocols have appeared on the market, such as Zigbee, Z-Wave, Wi-Fi, Bluetooth, etc. Each...
| |