Why is enumerate empty when passed a non-empty iterable? - python

It's my understanding that the enumerate() function will loop over each item in an iterable. However, in the example below is does not loop at all when given a non-empty iterable.
import itertools
x = itertools.chain(
itertools.combinations([1, 2, 3], 1),
itertools.combinations([1, 2, 3], 2),
itertools.combinations([1, 2, 3], 3),
)
print(list(x))
print(list(enumerate(x)))
print(list(enumerate(list(x))))
print(list(enumerate([(1, ), (2, ), (3, ), (1, 2), (1, 3), (2, 3), (1, 2, 3)])))
Output:
[(1,), (2,), (3,), (1, 2), (1, 3), (2, 3), (1, 2, 3)]
[]
[]
[(0, (1,)), (1, (2,)), (2, (3,)), (3, (1, 2)), (4, (1, 3)), (5, (2, 3)), (6, (1, 2, 3))]
Expected output:
[(1,), (2,), (3,), (1, 2), (1, 3), (2, 3), (1, 2, 3)]
[(0, (1,)), (1, (2,)), (2, (3,)), (3, (1, 2)), (4, (1, 3)), (5, (2, 3)), (6, (1, 2, 3))]
[(0, (1,)), (1, (2,)), (2, (3,)), (3, (1, 2)), (4, (1, 3)), (5, (2, 3)), (6, (1, 2, 3))]
[(0, (1,)), (1, (2,)), (2, (3,)), (3, (1, 2)), (4, (1, 3)), (5, (2, 3)), (6, (1, 2, 3))]
I'm running this with Python 3.8.2 on Arch Linux.
Can someone explain why enumerate() does not loop with these inputs?

You exhausted your chained iterator with the first print statement. The second and third did receive an empty iterator. If you want to see your expected output, you need to rebuild the chain for each run-through.
This is not specific to chain, nor to enumerate; you can get the same effect with
x = iter(range(10))

Related

How to select spesific list from nested lists with python [duplicate]

This question already has answers here:
Pythonic way of checking if a condition holds for any element of a list
(3 answers)
Apply function to each element of a list
(4 answers)
Closed 9 months ago.
Hello guys I have list like A
max_x=4
min_x=0
A=[[(0, 1), (1, 0), (1, 1), (2, 0)], [(0, 3), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 0), (4, 1), (4, 2)]]
A, includes different group of points (x,y) format.I wanted to find group if includes my max and min same time.Output should be like B.Because this cluster includes 0 and 4 as x.
B= [(0, 3), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 0), (4, 1), (4, 2)]
Thank you.
You could use a list comprehension to find any sublists of A that have a tuple that has x == min_x and also a tuple that has x == max_x:
max_x=4
min_x=0
A=[[(0, 1), (1, 0), (1, 1), (2, 0)], [(0, 3), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 0), (4, 1), (4, 2)]]
B = [l for l in A if any(x == min_x for x,_ in l) and any(x == max_x for x,_ in l)]
Output:
[[(0, 3), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 0), (4, 1), (4, 2)]]

Product of coordinates with specific order of iteration

Here's a snippet with a regular itertools.product usage:
from itertools import product
arr = [1,2,3]
pairs = list(product(arr, arr))
# pairs = [(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)]
Now I would like to have these points yielded in an order which can be achieved by sorting the resulting tuples in the following way:
sorted(pairs, key=lambda y:max(y))
# [(1, 1), (1, 2), (2, 1), (2, 2), (1, 3), (2, 3), (3, 1), (3, 2), (3, 3)]
Is there a way for me to input those numbers to itertools.product so it yields the tuples in this order, or do I need to sort the pairs after iterating over the results of itertools.product?
You can probably use your own modified approach to achieve this in one go.
sorted([(x, y) for x in [1, 2, 3] for y in [1, 2, 3]], key=lambda y:max(y))
OUTPUT
[(1, 1), (1, 2), (2, 1), (2, 2), (1, 3), (2, 3), (3, 1), (3, 2), (3, 3)]

find combinations in arbitrarily nested lists under conditions

I want to find possible paths on a finite grid of points. Say, starting point is (x,y). Then next point (m,n) in the path is given by conditions
(m!=x) and (n!=y) ie. I exclude the row and column I was in previously.
n < y ie. I always hop DOWN.
m,n >= 0 ie. all the points are always in first quadrant
Stopping criteria is when a point lies on x axis.
Hence, generate all possible combinations of such 'paths' possible.
Following is what I've tried.
def lisy(x,y):
return [(i,j) for i in range(4,0,-1) for j in range(4,0,-1) if(i!=x and j<y)]
def recurse(x,y):
if (not lisy(x,y)):
return (x,y)
else:
return [(x,y), [recurse(i,j) for i,j in lisy(x,y)]]
OUTPUT:
In [89]: recurse(1,4)
Out[89]:
[(1, 4),
[[(4, 3),
[[(3, 2), [(4, 1), (2, 1), (1, 1)]],
(3, 1),
[(2, 2), [(4, 1), (3, 1), (1, 1)]],
(2, 1),
[(1, 2), [(4, 1), (3, 1), (2, 1)]],
(1, 1)]],
[(4, 2), [(3, 1), (2, 1), (1, 1)]],
(4, 1),
[(3, 3),
[[(4, 2), [(3, 1), (2, 1), (1, 1)]],
(4, 1),
[(2, 2), [(4, 1), (3, 1), (1, 1)]],
(2, 1),
[(1, 2), [(4, 1), (3, 1), (2, 1)]],
(1, 1)]],
[(3, 2), [(4, 1), (2, 1), (1, 1)]],
(3, 1),
[(2, 3),
[[(4, 2), [(3, 1), (2, 1), (1, 1)]],
(4, 1),
[(3, 2), [(4, 1), (2, 1), (1, 1)]],
(3, 1),
[(1, 2), [(4, 1), (3, 1), (2, 1)]],
(1, 1)]],
[(2, 2), [(4, 1), (3, 1), (1, 1)]],
(2, 1)]]
This gives me a nested lists of possible new points from each point.
Can anyone tell me how to process my list obtained from recurse(1,4)?
edit1:
Effectively I hop from a given starting point (in a 4x4 grid [finite]), satisfying the three conditions mentioned until stopping criteria is met, ie. m,n > 0
I clarify the requirements I am working under in the docstring of my generator gridpaths(). Note that I have the horizontal size of the grid as a global variable and the vertical size of the grid is irrelevant, the x-coordinates of path points can be up to but not exceed that global value, and x-coordinates of non-consecutive path points can be equal (though consecutive path points must have different x-coordinates). I changed the name of the routine but kept the arguments as you had them. This version of my code adds the requirement that the y-coordinate of the final point on the path must be 1, and it also is safer in accepting arguments.
This is a generator of lists, so my test code shows how large the generator is then prints all the lists.
def gridpaths(x, y):
"""Generate all paths starting at (x,y) [x and y must be positive
integers] where, if (m,n) is the next point in the path after
(x,y), then m and n are positive integers, m <= xsize [xsize is a
global variable], m != x, and n < y, and so on for all consecutive
path points. The final point in the path must have a y-coordinate
of 1. Paths are yielded in lexicographic order."""
def allgridpaths(x, y, pathsofar):
"""Generate all such paths continuing from pathssofar without
the y == 1 requirement for the final path point."""
newpath = pathsofar + [(x, y)]
yield newpath
for m in range(1, xsize+1):
if m != x:
for n in range(1, y):
for path in allgridpaths(m, n, newpath):
yield path
x, y = max(int(x), 1), max(int(y), 1) # force positive integers
for path in allgridpaths(x, y, []):
# Only yield paths that end at y == 1
if path[-1][1] == 1:
yield path
# global variable: horizontal size of grid
xsize = 4
print(sum(1 for p in gridpaths(1, 4)), 'paths total.')
for p in gridpaths(1, 4):
print(p)
The printout shows that the point (1,4) in a 4x4 grid yields 48 paths. In fact, gridpaths(x, y) will return (xsize - 1) * xsize ** (y - 2) paths, which can grow very quickly. That is why I programmed a generator of lists rather than a list of lists. Let me know if your requirements are different from what I suppose. The printout from that code above is:
48 paths total.
[(1, 4), (2, 1)]
[(1, 4), (2, 2), (1, 1)]
[(1, 4), (2, 2), (3, 1)]
[(1, 4), (2, 2), (4, 1)]
[(1, 4), (2, 3), (1, 1)]
[(1, 4), (2, 3), (1, 2), (2, 1)]
[(1, 4), (2, 3), (1, 2), (3, 1)]
[(1, 4), (2, 3), (1, 2), (4, 1)]
[(1, 4), (2, 3), (3, 1)]
[(1, 4), (2, 3), (3, 2), (1, 1)]
[(1, 4), (2, 3), (3, 2), (2, 1)]
[(1, 4), (2, 3), (3, 2), (4, 1)]
[(1, 4), (2, 3), (4, 1)]
[(1, 4), (2, 3), (4, 2), (1, 1)]
[(1, 4), (2, 3), (4, 2), (2, 1)]
[(1, 4), (2, 3), (4, 2), (3, 1)]
[(1, 4), (3, 1)]
[(1, 4), (3, 2), (1, 1)]
[(1, 4), (3, 2), (2, 1)]
[(1, 4), (3, 2), (4, 1)]
[(1, 4), (3, 3), (1, 1)]
[(1, 4), (3, 3), (1, 2), (2, 1)]
[(1, 4), (3, 3), (1, 2), (3, 1)]
[(1, 4), (3, 3), (1, 2), (4, 1)]
[(1, 4), (3, 3), (2, 1)]
[(1, 4), (3, 3), (2, 2), (1, 1)]
[(1, 4), (3, 3), (2, 2), (3, 1)]
[(1, 4), (3, 3), (2, 2), (4, 1)]
[(1, 4), (3, 3), (4, 1)]
[(1, 4), (3, 3), (4, 2), (1, 1)]
[(1, 4), (3, 3), (4, 2), (2, 1)]
[(1, 4), (3, 3), (4, 2), (3, 1)]
[(1, 4), (4, 1)]
[(1, 4), (4, 2), (1, 1)]
[(1, 4), (4, 2), (2, 1)]
[(1, 4), (4, 2), (3, 1)]
[(1, 4), (4, 3), (1, 1)]
[(1, 4), (4, 3), (1, 2), (2, 1)]
[(1, 4), (4, 3), (1, 2), (3, 1)]
[(1, 4), (4, 3), (1, 2), (4, 1)]
[(1, 4), (4, 3), (2, 1)]
[(1, 4), (4, 3), (2, 2), (1, 1)]
[(1, 4), (4, 3), (2, 2), (3, 1)]
[(1, 4), (4, 3), (2, 2), (4, 1)]
[(1, 4), (4, 3), (3, 1)]
[(1, 4), (4, 3), (3, 2), (1, 1)]
[(1, 4), (4, 3), (3, 2), (2, 1)]
[(1, 4), (4, 3), (3, 2), (4, 1)]

Converting string that looks like a list into a real list - python

My input files have lines that looks like this:
[(0, 1), (1, 3), (2, 1), (3, 1), (4, 1)]
[(0, 1, 6), (1, 3,7), (3, 1,4), (3, 1,3), (8, 1,2)]
[1,2,3,5,3]
There are no letters, no decimals, only integers and number of element in tuples will be consistent.
How do i make them into real list of tuples / list of int?
Python comes with batteries included - that problem is solved by ast.literal_eval():
>>> import ast
>>> ast.literal_eval("[(0, 1), (1, 3), (2, 1), (3, 1), (4, 1)]")
[(0, 1), (1, 3), (2, 1), (3, 1), (4, 1)]
>>> ast.literal_eval("[(0, 1, 6), (1, 3,7), (3, 1,4), (3, 1,3), (8, 1,2)]")
[(0, 1, 6), (1, 3, 7), (3, 1, 4), (3, 1, 3), (8, 1, 2)]
>>> ast.literal_eval("[1,2,3,5,3]")
[1, 2, 3, 5, 3]

Making Combinations (Python)

In Python, is there a better way to get the set of combinations of n elements from a k-element set than nested for loops or list comprehensions?
For example, say from the set [1,2,3,4,5,6] I want to get [(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)]. Is there a better of of making it than
nums=[1,2,3,4,5,6]
doubles=[]
for a in nums:
for b in nums[a+1:]
doubles.append((a,b))
? It's okay if the elements of the list we end up with are sets, tuples, or lists; I just feel there should be an easier way to do this.
You can use itertools.combinations:
>>> from itertools import combinations
>>> nums = [1,2,3,4,5,6]
>>> list(combinations(nums, 2))
[(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6)]
The itertools module has a lot of really powerful tools that can be used in situations like this. In this case, you want itertools.combinations. Some other ones that you might find useful are itertools.combinations_with_replacement and itertools.permutations.
Example:
>>> import itertools
>>> list(itertools.combinations(range(1,7),2))
[(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6)]
>>> list(itertools.combinations_with_replacement(range(1,7),2))
[(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 5), (5, 6), (6, 6)]
>>> list(itertools.permutations(range(1,7),2))
[(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)]
You could use the itertools module
import itertools
alphabet = ['1','2','3','4','5','6']
combos = list(itertools.combinations(alphabet, 2))
print combos

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