CODEWITHSUNDEEP
pythonlist
zelenov aleksey
zelenov aleksey

Reputation: 398

how to check list before adding items?

I have a list of [i,j] values:

lst = [[1,2],[1,3],[3,4],[4,6]]
#[1,2] means that i=1 and j=2

and add new elenemt [a,b], for example [6,1], how can I add values to my lst by condition

(add element [a,b] only if [b,j] does not exists in a set)

So element [6,1] should not be added, because [1,2] and [1,3] is already in the list

Thanks in advance for any help!

Upvotes: 0

Views: 846

Answers (2)

bruno desthuilliers
bruno desthuilliers

Reputation: 77912

From a readability / maintainability POV I usually prefer to have a custom class handling the invariants instead:

# XXX poor naming but I don't know enough of the context
class Container(object): 
    def __init__(self, *initvals):
        self._values = []
        self._keys = set()
        for pair in initvals:
            self.add(pair)

    def add(self, pair):
        if pair[1] in self._keys:
            raise ValueError("second item of pair '{}' already set".format(pair))
        self._keys.add(pair[0])
        self._values.append(pair)

    def values(self):
        # returns a copy so no one accidentally messes
        # with internal state
        return self._values[:]

c = Container((1,2), (1,3), (3,4), (4,6))
c.add((8, 9))
assert (8, 9) in c.values(), "(8,9) not correctly added"

try:
    c.add((6, 1))
except ValueError as e:
    #print "got error {} when adding (6,1)".format(e)
    pass
else:
    assert False, "adding (6, 1) should have raised a ValueError"

print c.values()

Upvotes: 2

tobias_k
tobias_k

Reputation: 82899

You could just iterate the list and check whether for any of the items (i,j) in the list, i is the same as b in the item (a,b) you want to add. (At least that's how I understood your condition, but if the condition is slightly different, this should be easy to adapt accordingly.)

def cond_add(lst, item):
    a, b = item
    if not any(i == b for (i, j) in lst):
        lst.append(item)

lst = [[1,2],[1,3],[3,4],[4,6]]
cond_add(lst, [6,1])
cond_add(lst, [6,5])
print(lst)

If you want to add multiple elements, it would pay to create a set from the first elements, so you can do the check in constant time for each individual item. (You could do the same for adding individual elements, but you'd have to update the set after each item.)

def cond_add_many(lst, items):
    first = set(i for (i, j) in lst)
    for (a,b) in items:
        if b not in first:
            lst.append((a,b))
            first.add(a)

lst = [[1,2],[1,3],[3,4],[4,6]]
cond_add_many(lst, [[6,1], [6,5], [7,6]])
print(lst)

Afterwards, in both cases, lst is [[1, 2], [1, 3], [3, 4], [4, 6], (6, 5)]

Upvotes: 2

Related Questions