CODEWITHSUNDEEP
pythonlist-comprehension
John Lexus
John Lexus

Reputation: 3626

Can you break this pythonic list comprehension down

I encountered this List Comprehension when looking for interesting ways to solve longest increasing subsequence problems:

def lis(a):
    L = []
    for (k,v) in enumerate(a):
        L.append(max([L[i] for (i,n) in enumerate(a[:k]) if n<v] or [[]], key=len) + [v])
    return max(L, key=len)

I've tried to understand it, but this is the best way I could understand it:

def lis(a):
    L = []
    for (k,v) in enumerate(a):
        for (i,n) in enumerate(a[:k]):
            if n < v:
                L.append(max(L[i]) + [v])
            else:
                L.append([[]])
        #L.append(max([L[i] for (i,n) in enumerate(a[:k]) if n<v] or [[]], key=len) + [v])
    return max(L, key=len)

But this is incorrect... any help? Thank you.

Edit:

New code:

def lis(a):
    L = []
    for (k,v) in enumerate(a):
        for (i,n) in enumerate(a[:k]):
            if n < v:
                L.append(max(L[i], key=len) + [v])
            else:
                L.append([[]] + [v])

When I run this on this list:

a = [5, 2, 1, 5, 3, 2, 10, 2, 4, 2, 1, 0, 5]

I get this error:

  L.append(max(L[i], key=len) + [v])
TypeError: object of type 'int' has no len()

Upvotes: 2

Views: 77

Answers (1)

ence ladus
ence ladus

Reputation: 62

You forgot + [v] in the else statement.

Edit (think it's right):

def lis(a):
    L = []
    for (k, v) in enumerate(a):
        tmp = []
        for (i, n) in enumerate(a[:k]):
            if n < v:
                tmp.append(L[i])
        if len(tmp) == 0:
            tmp = [[]]
        L.append(max(tmp, key=len) + [v])
    return max(L, key=len)

Upvotes: 2

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