CODEWITHSUNDEEP
pythonpython-3.x
Andrew
Andrew

Reputation: 431

min() arg is an empty sequence

I'm trying to find minimum element in matrix row, but there are two conditions: 1) it must be > 0 2) and this point must be not visited(is_visited[k] is False) I'm trying to do next:

min(x for x in matr_sum[i] if x > 0  if is_visited[k] is False )

But there is an error: min() arg is an empty sequence

The full block of code:

for k in range(4):
        if matr_sum[i][k] == min(x for x in matr_sum[i] if x > 0  if is_visited[k] is False ) and i!=k:
            return k

How to resolve it? Or should I write my min() function? Because it works with one condition:

min(x for x in matr_sum[i] if x > 0)

But with two conditions, it doesn't work.

Upvotes: 33

Views: 142437

Answers (4)

patrick
patrick

Reputation: 4852

If you want to avoid this ValueError in general, you can set a default argument to min(), that will be returned in case of an empty list. See described here.

min([], default="EMPTY")
# returns EMPTY

Note that this only works in Python 3.4+

Upvotes: 49

Steve Jessop
Steve Jessop

Reputation: 279255

There is no problem with the syntax. It's certainly unusual to have two if clauses, but it's allowed. Consider:

print(min(x for x in range(1,300) if x % 3 == 0 if x % 5 == 0))

Output:

15

However:

print(min(x for x in range(1,300) if x % 2 != 0 if x % 2 != 1))

Output:

ValueError: min() arg is an empty sequence

There are no integers that are both odd and even, so there are no values for min to see, so it throws an exception.

I deduce that in your code, there are no values that pass both conditions. Python doesn't allow you to compute "the minimum of no values", mainly because it makes no sense.

You have to decide what you want to do in the case where there is no minimum because there are no values greater than 0. For example, if you don't want to return k in that case then I might re-write your code something like this:

for k in range(4):
    if k != i and is_visited[k] is False:
        if matr_sum[i][k] > 0 and matr_sum[i][k] == min(x for x in matr_sum[i] if x > 0):
            return k

It might not be obvious why this helps, but assuming matr_sum[i] is a list or similar, then once we know matr_sum[i][k] > 0 then we know the generator passed to min isn't empty, so we won't get an exception. Whereas if matr_sum[i][k] <= 0, then it certainly isn't equal to the smallest positive value, so there's no need to compute the min at all. Another way to write that would be:

if matr_sum[i][k] > 0 and not any(0 < x < matr_sum[i][k] for x in matr_sum[i])

Actually, I'd normally write if not is_visited[k], but I leave it as is False since I don't know whether changing it would change the behaviour of your code.

Upvotes: 11

Tom Dalton
Tom Dalton

Reputation: 6190

Try this - it creates the list of x values xs and then only tries to find the min if xs is non-empty. You may need to add some logic to handle the case that xs is empty, depending on what your code is doing.

for k in range(4):
    if is_visited[k] is False and i != k:
        xs = [x for x in matr_sum[i] if x > 0]
        if xs and matr_sum[i][k] == min(xs):
            return k

Upvotes: 2

Kasravnd
Kasravnd

Reputation: 107287

Just use and operation for concatenate tow if statement :

min(x for x in matr_sum[i] if x > 0 and if is_visited[k] is False and i!=k)

Upvotes: 1

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