I'm committing a crime in python on purpose. This is a bad way of doing this.
The GOAL here is cursed code. All on one line.
I have basically whats below
with open("file") as f:
[int(x) for x in [y for y in f.read().split()]
I cannot use
with open("file") as f:
a = f.read().split()
[x for x in [(a[i-1, e]) for i, e in enumerate(a) if i > 0] ...]
because the goal is to have this in one line (aside from the with open)
I would like to return from the original object the current element and either the previous one or the next one.
To illustrate it clearly.
a = [1, 2, 3, 4, 5]
After the illegal code would return
[(1, 2), (2, 3), (3, 4), (4, 5), (5, ?)]
So again the focus here is not production code. This is purely to see how much we can abuse the language.
So far I've found https://code.activestate.com/recipes/204297/ which references the use of local in python2, after mucking around with it I found that the interface for it is a little different.
I've been able to get the object in memory but I dont know how to actually use this object now that I have it.
local()['.0']
Most attributes seem to be missing, no __self__ to call.
Please share your most cursed ideas for this.
Normally, I would use tee and islice on the generator for something like this:
from itertools import tee, islice
with open("file") as f:
a, b = tee(f.read().split())
b = islice(b, 1, None)
list(zip(a, b))
You can convert this into a one-liner using (abusing) the walrus operator (:=):
list(zip((k := tee(f.read().split()))[0], islice(k[1], 1, None)))
The result is
[(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)]
If you want the last element to be padded, use zip_longest instead of zip:
from itertools import tee, islice, zip_longest
...
list(zip_longest((k := tee(f.read().split()))[0], islice(k[1], 1, None), fillvalue='?'))
The result in this case is
[(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, '?')]
The nice thing about using iterators rather than lists this way is that while f.read().split() is a sequence of known length, the tee and islice will work on any iterable, even if the length is unknown.
I am a newbee in python and programing, I am trying to come up with combinations and weed out combinations with certain conditions.
So in the case below, I have tried to generate all possible combinations between 1-100. But I don't know where to go after this.
import itertools
i_list = []
for i in range (1, 101):
i_list.append(i)
comb = itertools.combinations(i_list,2)
for combinations in list(comb):
print (combinations)
This runs fine and will generate a list from 1-100, and give me an output of
(1,2) (1,3).........(98,99) (98,100) (99,100)
Now my goal is to weed out the combinations with a difference < 5, so for example: (1,2) the difference is less than 5, so it should not be outputted. (1,8) the difference is greater than 5, so it should be outputted. I hope that make sense.
Can anyone guide me through the thought process and suggest an easy approach?
You can use itertools.filterfalse for this and then iterate over the result.
Also, with iterators, you want to wait until you really need a list before you convert to a list with list(). There's no reason to ever do that in this case because you are always iterating. This allows you to work with very large sets without taking up the memory and time of running through the iterator just to make a list to then iterate the list:
from itertools import combinations, filterfalse
comb = combinations(range(1, 101),2)
filtered = filterfalse(lambda x: abs(x[0] - x[1]) < 5, comb)
for combinations in filtered:
print (combinations)
The iterators produced by range(), combinations and fitleredfalse are all lazy, so they never start evaluating until you start looping over them. This allows you to defer any work until it needs to be done or to iterate over part of a large set without calculating the entire thing.
You can use a list comprehension to restrict the generated values to be kept inside the list:
from itertools import combinations
comb = [ x for x in combinations(range(1,101),2) if x[1]-x[0]>4 ]
print (comb)
Output:
[(1, 6), (1, 7), (1, 8), ... snipp ..., (93, 99), (93, 100), (94, 99), (94, 100), (95, 100)]
combinations respects the order of numbers so no abs() around x[1]-x[0] needed - range itself is a sequence and your resulting list weeds out all numbers you do not want due to the if x[1]-x[0]>4 condition.
This should accomplish what you are asking:
>>> import itertools
>>> combinations = itertools.combinations(range(1, 101), 2)
>>> generator = ((a, b) for a, b in combinations if b - a >= 5)
>>> for pair in generator:
print(pair, end=' ')
(1, 6) (1, 7) (1, 8) (1, 9) (1, 10) (1, 11) (1, 12) (1, 13) (1, 14) (1, 15) ...
Alternatively, you can try this instead to do the exact same thing:
>>> generator = ((a, b) for a in range(1, 96) for b in range(a + 5, 101))
>>> for pair in generator:
print(pair, end=' ')
(1, 6) (1, 7) (1, 8) (1, 9) (1, 10) (1, 11) (1, 12) (1, 13) (1, 14) (1, 15) ...
Is there a nice Pythonic way to loop over a list, retuning a pair of elements? The last element should be paired with the first.
So for instance, if I have the list [1, 2, 3], I would like to get the following pairs:
1 - 2
2 - 3
3 - 1
A Pythonic way to access a list pairwise is: zip(L, L[1:]). To connect the last item to the first one:
>>> L = [1, 2, 3]
>>> zip(L, L[1:] + L[:1])
[(1, 2), (2, 3), (3, 1)]
I would use a deque with zip to achieve this.
>>> from collections import deque
>>>
>>> l = [1,2,3]
>>> d = deque(l)
>>> d.rotate(-1)
>>> zip(l, d)
[(1, 2), (2, 3), (3, 1)]
I'd use a slight modification to the pairwise recipe from the itertools documentation:
def pairwise_circle(iterable):
"s -> (s0,s1), (s1,s2), (s2, s3), ... (s<last>,s0)"
a, b = itertools.tee(iterable)
first_value = next(b, None)
return itertools.zip_longest(a, b,fillvalue=first_value)
This will simply keep a reference to the first value and when the second iterator is exhausted, zip_longest will fill the last place with the first value.
(Also note that it works with iterators like generators as well as iterables like lists/tuples.)
Note that #Barry's solution is very similar to this but a bit easier to understand in my opinion and easier to extend beyond one element.
I would pair itertools.cycle with zip:
import itertools
def circular_pairwise(l):
second = itertools.cycle(l)
next(second)
return zip(l, second)
cycle returns an iterable that yields the values of its argument in order, looping from the last value to the first.
We skip the first value, so it starts at position 1 (rather than 0).
Next, we zip it with the original, unmutated list. zip is good, because it stops when any of its argument iterables are exhausted.
Doing it this way avoids the creation of any intermediate lists: cycle holds a reference to the original, but doesn't copy it. zip operates in the same way.
It's important to note that this will break if the input is an iterator, such as a file, (or a map or zip in python-3), as advancing in one place (through next(second)) will automatically advance the iterator in all the others. This is easily solved using itertools.tee, which produces two independently operating iterators over the original iterable:
def circular_pairwise(it):
first, snd = itertools.tee(it)
second = itertools.cycle(snd)
next(second)
return zip(first, second)
tee can use large amounts of additional storage, for example, if one of the returned iterators is used up before the other is touched, but as we only ever have one step difference, the additional storage is minimal.
There are more efficient ways (that don't built temporary lists), but I think this is the most concise:
> l = [1,2,3]
> zip(l, (l+l)[1:])
[(1, 2), (2, 3), (3, 1)]
Pairwise circular Python 'for' loop
If you like the accepted answer,
zip(L, L[1:] + L[:1])
you can go much more memory light with semantically the same code using itertools:
from itertools import islice, chain #, izip as zip # uncomment if Python 2
And this barely materializes anything in memory beyond the original list (assuming the list is relatively large):
zip(l, chain(islice(l, 1, None), islice(l, None, 1)))
To use, just consume (for example, with a list):
>>> list(zip(l, chain(islice(l, 1, None), islice(l, None, 1))))
[(1, 2), (2, 3), (3, 1)]
This can be made extensible to any width:
def cyclical_window(l, width=2):
return zip(*[chain(islice(l, i, None), islice(l, None, i)) for i in range(width)])
and usage:
>>> l = [1, 2, 3, 4, 5]
>>> cyclical_window(l)
<itertools.izip object at 0x112E7D28>
>>> list(cyclical_window(l))
[(1, 2), (2, 3), (3, 4), (4, 5), (5, 1)]
>>> list(cyclical_window(l, 4))
[(1, 2, 3, 4), (2, 3, 4, 5), (3, 4, 5, 1), (4, 5, 1, 2), (5, 1, 2, 3)]
Unlimited generation with itertools.tee with cycle
You can also use tee to avoid making a redundant cycle object:
from itertools import cycle, tee
ic1, ic2 = tee(cycle(l))
next(ic2) # must still queue up the next item
and now:
>>> [(next(ic1), next(ic2)) for _ in range(10)]
[(1, 2), (2, 3), (3, 1), (1, 2), (2, 3), (3, 1), (1, 2), (2, 3), (3, 1), (1, 2)]
This is incredibly efficient, an expected usage of iter with next, and elegant usage of cycle, tee, and zip.
Don't pass cycle directly to list unless you have saved your work and have time for your computer to creep to a halt as you max out its memory - if you're lucky, after a while your OS will kill the process before it crashes your computer.
Pure Python Builtin Functions
Finally, no standard lib imports, but this only works for up to the length of original list (IndexError otherwise.)
>>> [(l[i], l[i - len(l) + 1]) for i in range(len(l))]
[(1, 2), (2, 3), (3, 1)]
You can continue this with modulo:
>>> len_l = len(l)
>>> [(l[i % len_l], l[(i + 1) % len_l]) for i in range(10)]
[(1, 2), (2, 3), (3, 1), (1, 2), (2, 3), (3, 1), (1, 2), (2, 3), (3, 1), (1, 2)]
I would use a list comprehension, and take advantage of the fact that l[-1] is the last element.
>>> l = [1,2,3]
>>> [(l[i-1],l[i]) for i in range(len(l))]
[(3, 1), (1, 2), (2, 3)]
You don't need a temporary list that way.
Amazing how many different ways there are to solve this problem.
Here's one more. You can use the pairwise recipe but instead of zipping with b, chain it with the first element that you already popped off. Don't need to cycle when we just need a single extra value:
from itertools import chain, izip, tee
def pairwise_circle(iterable):
a, b = tee(iterable)
first = next(b, None)
return izip(a, chain(b, (first,)))
I like a solution that does not modify the original list and does not copy the list to temporary storage:
def circular(a_list):
for index in range(len(a_list) - 1):
yield a_list[index], a_list[index + 1]
yield a_list[-1], a_list[0]
for x in circular([1, 2, 3]):
print x
Output:
(1, 2)
(2, 3)
(3, 1)
I can imagine this being used on some very large in-memory data.
This one will work even if the list l has consumed most of the system's memory. (If something guarantees this case to be impossible, then zip as posted by chepner is fine)
l.append( l[0] )
for i in range( len(l)-1):
pair = l[i],l[i+1]
# stuff involving pair
del l[-1]
or more generalizably (works for any offset n i.e. l[ (i+n)%len(l) ] )
for i in range( len(l)):
pair = l[i], l[ (i+1)%len(l) ]
# stuff
provided you are on a system with decently fast modulo division (i.e. not some pea-brained embedded system).
There seems to be a often-held belief that indexing a list with an integer subscript is un-pythonic and best avoided. Why?
This is my solution, and it looks Pythonic enough to me:
l = [1,2,3]
for n,v in enumerate(l):
try:
print(v,l[n+1])
except IndexError:
print(v,l[0])
prints:
1 2
2 3
3 1
The generator function version:
def f(iterable):
for n,v in enumerate(iterable):
try:
yield(v,iterable[n+1])
except IndexError:
yield(v,iterable[0])
>>> list(f([1,2,3]))
[(1, 2), (2, 3), (3, 1)]
How about this?
li = li+[li[0]]
pairwise = [(li[i],li[i+1]) for i in range(len(li)-1)]
from itertools import izip, chain, islice
itr = izip(l, chain(islice(l, 1, None), islice(l, 1)))
(As above with #j-f-sebastian's "zip" answer, but using itertools.)
NB: EDITED given helpful nudge from #200_success. previously was:
itr = izip(l, chain(l[1:], l[:1]))
If you don't want to consume too much memory, you can try my solution:
[(l[i], l[(i+1) % len(l)]) for i, v in enumerate(l)]
It's a little slower, but consume less memory.
Starting in Python 3.10, the new pairwise function provides a way to create sliding pairs of consecutive elements:
from itertools import pairwise
# l = [1, 2, 3]
list(pairwise(l + l[:1]))
# [(1, 2), (2, 3), (3, 1)]
or simply pairwise(l + l[:1]) if you don't need the result as a list.
Note that we pairwise on the list appended with its head (l + l[:1]) so that rolling pairs are circular (i.e. so that we also include the (3, 1) pair):
list(pairwise(l)) # [(1, 2), (2, 3)]
l + l[:1] # [1, 2, 3, 1]
Just another try
>>> L = [1,2,3]
>>> zip(L,L[1:]) + [(L[-1],L[0])]
[(1, 2), (2, 3), (3, 1)]
L = [1, 2, 3]
a = zip(L, L[1:]+L[:1])
for i in a:
b = list(i)
print b
this seems like combinations would do the job.
from itertools import combinations
x=combinations([1,2,3],2)
this would yield a generator. this can then be iterated over as such
for i in x:
print i
the results would look something like
(1, 2)
(1, 3)
(2, 3)
I know that izip can read two files at a time, but I don't know how to iterate over the two files at the same time (or if it is rational to do that). Here is my first step on the code that should read the docs line by line and return whether the docs have the same amount of lines:
from sys import argv
import itertools
a = 0
b = 0
doc1 = open(argv[1],"r")
doc2 = open(argv[2],"r")
for lineA,lineB in itertools.izip(doc1,doc2):
a = a + 1
b = b + 1
if a/b == 1:
print "equal number of lines in docs"
else:
print "docs with different number of lines"
Here's a simple example to demonstrate why that won't work:
>>> a = range(5)
>>> b = range(10)
>>> zip(a, b)
[(0, 0), (1, 1), (2, 2), (3, 3), (4, 4)] # ...is that it?
zip and itertools.izip truncate whichever iterable is longest, so you'll always conclude that the two files are the same length.
Instead, use itertools.izip_longest, which won't truncate the longer file:
>>> from itertools import izip_longest
>>> list(izip_longest(a, b))
[(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), (None, 5), (None, 6), (None, 7), (None, 8), (None, 9)]
# ^ default fillvalue
Now you can check e.g. if lineA is None: to detect when a file has run out.
Alternatively, you can count the number of '\n' like this:
from sys import argv
with open(argv[1],"r") as f1, open(argv[2],"r") as f2:
if f1.read().count('\n') == f2.read().count('\n'):
print "equal number of lines in docs"
else:
print "docs with different number of lines"