I wanted to find a better way to loop through orthodiagonal indices in order, I am currently using numpy but I think I'm making an unnecessary number of function calls.
import numpy as np
len_x, len_y = 50, 50 #they don't have to equal
index_arr = np.add.outer(np.arange(len_x), np.arange(len_y))
Currently, I am looping through like this:
for i in range(np.max(index_arr)):
orthodiag_indices = zip(*np.where(index_arr == i))
for index in orthodiag_indices:
# DO FUNCTION OF index #
I have an arbitrary function of the index tuple, index and other parameters outside of this loop. It feels like I don't need the second for loop, and I should be able to do the whole thing in one loop. On top of this, I'm making a lot of function calls from zip(*np.where(index_arr == i)) for every i. What's the most efficient way to do this?
Edit: should mention that it's important that the function applies to index_arr == i in order, i.e., it does 0 first, then 1, then 2 etc. (the order of the second loop doesn't matter).
Edit 2: I guess what I want is a way to get the indices [(0,0), (0,1), (1,0), (2,0), (1,1), (2,0), ...] efficiently. I don't think I can apply a vectorized function because I am populating an np.zeros((len_x, len_y)) array, and going back to the first edit, the order matters.
You could use tril/triu_indices. Since the order of the (former) inner loop doesn't matter dimensions can be swapped as needed, I'll assume L>=S:
L,S = 4,3
a0,a1 = np.tril_indices(L,0,S)
b0,b1 = np.triu_indices(S,1)
C0 = np.concatenate([a0-a1,b0+L-b1])
C1 = np.concatenate([a1,b1])
*zip(C0,C1),
# ((0, 0), (1, 0), (0, 1), (2, 0), (1, 1), (0, 2), (3, 0), (2, 1), (1, 2), (3, 1), (2, 2), (3, 2))
I think itertools.product() will be of use here
import itertools as it
x,y = 2,3
a=list(it.product(range(x),range(y))
which gives a as
[(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)]
If you need them in order then,
b=np.argsort(np.sum(a,1))
np.array(a)[b]
which gives,
array([[0, 0],
[0, 1],
[1, 0],
[0, 2],
[1, 1],
[1, 2]])
Hope that helps!
Related
Having this frustrating issue where i want to pass through the tuples in the following list
through a method on another list of instances of a class that i have created
list_1=[(0, 20), (10, 1), (0, 1), (0, 10), (5, 5), (10, 50)]
instances=[instance[0], instance[1],...instance[n]]
results=[]
pos_list=[]
for i in range(len(list_1)):
a,b=List_1[i]
result=sum(instance.method(a,b) for instance in instances)
results.append(result)
if result>=0:
pos_list.append((a,b))
print(results)
print(pos_list)
the issue is that all instances are taking the same tuple, where as i want the method on the first instance to take the first tuple and so on.
I ultimately want to see it append to the new list (pos_list) if the sum is >0.
Anyone know how i can iterate this properly?
EDIT
It will make it clearer if I print the result of the sum also.
Basically I want the sum to perform as follows:
result = instance[0].method(0,20), instance[1].method(10,1), instance[2].method(0,1), instance[3].method(0,10), instance[4].method(5,5), instance[5].method(10,50)
For info the method is just the +/- product of the two values depending on the attributes of the instance.
So results for above would be:
result = [0*20 - 10*1 - 0*1 + 0*10 - 5*5 + 10*50] = [465]
pos_list=[(0, 20), (10, 1), (0, 1), (0, 10), (5, 5), (10, 50)]
except what is actually doing is using the same tuple for all instances like this:
result = instance[0].method(0,20), instance[1].method(0,20), instance[2].method(0,20), instance[3].method(0,20), instance[4].method(0,20), instance[5].method(0,20)
result = [0*20 - 0*20 - 0*20 + 0*20 - 0*20 + 0*20] = [0]
pos_list=[]
and so on for (10,1) etc.
How do I make it work like the first example?
You can compute your sum using zip to generate all the pairs of correspondent instances and tuples.
result=sum(instance.payout(*t) for instance, t in zip(instances, List_1))
The zip will stop as soon as it reaches the end of the shortest of the two iterators. So if you have 10 instances and 100 tuples, zip will produce only 10 pairs, using the first 10 elements of both lists.
The problem I see in your code is that you are computing this sum for each element of List_1, so if payout produces always the same result with the same inputs (e.g., it has no memory or randomness), the value of result will be the same at each iteration. So, in the end, results will be composed by the same value repeated a number of times equal to the length of List_1, while pos_list will contain all (the sum is greater than 0) or none (the sum is less or equal to zero) of the input tuples.
Instead, it would make sense if items of List_1 were lists or tuples themselves:
List_1 = [
[(0, 1), (2, 3), (4, 5)],
[(6, 7), (8, 9), (10, 11)],
[(12, 13), (14, 15), (16, 17)],
]
So, in this case, supposing that your class for instances is something like this:
class Goofy:
def __init__(self, positive_sum=True):
self.positive_sum = positive_sum
def payout(self, *args):
if self.positive_sum:
return sum(args)
else:
return -1 * sum(args)
instances = [Goofy(i) for i in [True, True, False]]
you can rewrite your code in this way:
results=[]
pos_list=[]
for el in List_1:
result = sum(g.payout(*t) for g, t in zip(instances, el))
results.append(result)
if result >= 0:
pos_list.append(el)
Running the previous code, results will be:
[-3, 9, 21]
while pop_list:
[[(6, 7), (8, 9), (10, 11)], [(12, 13), (14, 15), (16, 17)]]
If you are interested only in pop_list, you can compact your code in only one line:
pop_list = list(filter(lambda el: sum(g.payout(*t) for g, t in zip(instances, el)) > 0, List_1))
many thanks for the above! I have it working now.
Wasn't able to use args given my method had a bit more to it but the use of zip is what made it click
import random
rand=random.choices(list_1, k=len(instances))
results=[]
pos_list=[]
for r in rand:
x,y=r
result=sum(instance.method(x,y) for instance,(x,y) in zip(instances, rand))
results.append(result)
if result>=0:
pos_list.append(rand)
print(results)
print(pos_list)
for list of e.g.
rand=[(20, 5), (0, 2), (0, 100), (2, 50), (5, 10), (50, 100)]
this returns the following
results=[147]
pos_list=[(20, 5), (0, 2), (0, 100), (2, 50), (5, 10), (50, 100)]
so exactly what I wanted. Thanks again!
I am looking for a way to find the biggest 5 weighted edges in a node. Is there a way to specify that I want exactly the biggest 5 edges without a specific threshold value(a.k.a universal for any weighted graph)?
You could consider the edges sorted by weight and build a dictionary that maps a node with its edges, sorted by weight in a non-increasing way.
>>> from collections import defaultdict
>>> res = defaultdict(list)
>>> for u,v in sorted(G.edges(), key=lambda x: G.get_edge_data(x[0], x[1])["weight"], reverse=True):
... res[u].append((u,v))
... res[v].append((u,v))
...
Then, given a node (e.g., 0), you could get the top N (e.g., 5) weighted edges as
>>> res[0][:5]
[(0, 7), (0, 2), (0, 6), (0, 1), (0, 3)]
If you only need to do it for a node (e.g., 0), you can directly do:
>>> sorted_edges_u = sorted(G.edges(0), key=lambda x: G.get_edge_data(x[0], x[1])["weight"], reverse=True)
>>> sorted_edges_u[:5]
[(0, 7), (0, 2), (0, 6), (0, 1), (0, 3)]
I have 2 list where each list is of size 250000. I wanted to iterate thru the lists and return the values that are greater than 3.
For example:
import itertools
from array import array
import numpy as np
input = (np.array([list([8,1]), list([2,3,4]), list([5,3])],dtype=object), np.array([1,0,0,0,1,1,1]))
X = input[0]
y = input[1]
res = [ u for s in X for u in zip(y,s) ]
res
I don't get the expected output.
Actual res : [(1, 8), (0, 1), (1, 2), (0, 3), (0, 4), (1, 5), (0, 3)]
Expected output 1 : [(8,1), (1,0), (2, 0), (3, 0), (4, 1), (5, 1), (3, 1)]
Expected output 2 : [(8,1), (4, 1), (5, 1))] ---> for greater than 3
I took references from stackoverflow. Tried itertools as well.
Using NumPy to store lists of non-uniform lengths creates a whole lot of issues, like the ones you are seeing. If it were an array integers, you could simply do
X[X > 3]
but since it is an array of lists, you have to jump through all sorts of hoops to get what you want, and basically lose all the advantages of using NumPy in the first place. You could just as well use lists of lists and skip NumPy altogether.
As an alternative I would recommend using Pandas or something else more suitable than NumPy:
import pandas as pd
df = pd.DataFrame({
'group': [0, 0, 1, 1, 1, 2, 2],
'data': [8, 1, 2, 3, 4, 5, 4],
'flag': [1, 0, 0, 0, 1, 1, 1],
})
df[df['data'] > 3]
# group data flag
# 0 0 8 1
# 4 1 4 1
# 5 2 5 1
# 6 2 4 1
Use filter
For example:
input = [1, 3, 2, 5, 6, 7, 8, 22]
# result contains even numbers of the list
result = filter(lambda x: x % 2 == 0, input)
This should give you result = [2, 6, 8, 22]
Not sureI quite understand exactly what you're trying to do... but filter is probably a good way.
Given a list of tuples, [(x, y, z), ....., (x_n, y_n,z_n)], x, y are nonnegative number and z is either 0 or 1, I want to sort the list based on the following three criteria-
if x_i != x_j, sort on ascendening order of x(tuple[0])
if x_i == x_j and z_i != z_j, sort on ascendening order of z(tuple[2])
if x_i == x_j and z_i == z_j and z_i == 0, sort on descending order of y(tuple[1])
if x_i == x_j and z_i == z_j and z_i == 1, sort on ascending order of y(tuple[1])
Input: [(1, 1, 0), (2, 1, 1), (1, 2, 0), (2, 2, 1), (1, 3, 0), (2, 3, 1)]
output:[(1, 3, 0), (1, 2, 0), (1, 1, 0), (2, 1, 1), (2, 2, 1), (2, 3, 1)]
Since Python 3 does not support custom comparator function for sort as I know for JAVA, I do not know how to incorporate the above three criteria in the sort method.
I can sort based on the two criteria (either 1,2 or 1,3) of the above-mentioned criterion. Adding the third criteria makes one of 2 or 3 invalid. I am adding my code here-
points.sort(key=lambda p: p[2])
points.sort(key=lambda p: p[1], reverse=True)
points.sort(key=lambda p: p[0])
OUTPUT: [(1, 3, 0), (1, 2, 0), (1, 1, 0), (2, 3, 1), (2, 2, 1), (2, 1, 1)] (criteria 3 not satisfied)
Can anybody suggest, what should be the value of key argument in this situation? Thanks
Just encoding your criteria...
points.sort(key=lambda p: (p[0], p[2], p[1] if p[2] else -p[1]))
If you have truly ridiculously complicated sorting rules, you can just write a comparator function, then use functools.cmp_to_key to make it into a valid key argument. So write your insane comparator function, add from functools import cmp_to_key to the top of your file, then do:
points.sort(key=cmp_to_key(my_insane_comparator))
and it will work as expected. All cmp_to_key really does is make a custom class with a custom __lt__ (less than operator) that performs the work of the comparator in the __lt__ on each comparison.
I can't figure out how to get permutations to return the actual permutation and not
I tried a lot of different things to no avail. The code I used was from itertools import permutations and then permutations([1,2,3]). Thanks!
This may not be answering your question (it appears to be missing the part after 'and not'), but from your code, what you are likely seeing is the repr of the itertools.permutations iterator. You can iterate through this object just as you would a normal list in order to access all of the items. If you want to convert it to a list, you can wrap it in list:
>>> from itertools import permutations
>>> permutations([1, 2, 3])
<itertools.permutations object at 0x1e67890>
>>> list(permutations([1, 2, 3]))
[(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)]
However as mentioned above, the iterator can be iterated over just like you would a normal list (the benefit of returning an iterator is that the entire sequence is not loaded into memory right away - it is instead loaded 'as needed'):
>>> for perm in permutations([1, 2, 3]):
... print(perm)
...
(1, 2, 3)
(1, 3, 2)
(2, 1, 3)
(2, 3, 1)
(3, 1, 2)
(3, 2, 1)
itertools.permutations is a generator, which means you have to retrieve results from it by using it like this:
for permutation in itertools.permutations([1,2,3]):
do_stuff_with(permutation)
or alternatively put all of them in a list:
list(itertools.permutations([1,2,3]))
or, less conveniently:
generator = itertools.permutations([1,2,3])
generator.__next__()
from itertools import permutations
#iteration
for p in permutations([1,2,3]):
print(p)
This should work perfectly.