CODEWITHSUNDEEP
pythonlist
Stupidyeti
Stupidyeti

Reputation: 55

Python - Searching index of lists in list containing one element

I have a list L of 4-length list

L = [[1,2,12,13],[2,3,13,14],...]

and two integers a and b which appear many times in the sublists. What I want is to find the index of the sublists in L which contain a AND b.

I wrote a little code

l=[]
for i in range(len(L)):
    if L[i][0]==a or L[i][1]==a or L[i][2]==a or L[i][3]==a:
        l.append([i] + L[i]) # I put the index in the first position.
# Now l is a list of 5-length lists.
# I do the same loop on that list.
r=[]
for i in range(len(l)):
    if l[i][1]==b or l[i][2]==b or l[i][3]==b or l[i][4]==b:
        r.append(i)

The index I am looking for are in the list r. However I am pretty sure there is another way to do it in Python since I barely know this language. Maybe if my variable L is something else than a list of lists it would be easier/faster, because I will call this procedure a lot in my main program. (len(L) is around 3000)

By the way I know that the number of index is between one and four included, so I could put some break but I don't know if it will be faster.

---------------- EDIT 1 ----------------

Change "a or b (or is inclusive)" to "a AND b" in the second sentence. I wrote a mistake about my goal.

Upvotes: 0

Views: 103

Answers (4)

user2314737
user2314737

Reputation: 29307

Try with

for index, item in enumerate(L):
  if a in item or b in item:
    r.append(index)

Upvotes: 0

gabe
gabe

Reputation: 2511

This kind of thing is what list comprehension is made for.

If you really want inclusive or -- then this is the list you want. In your code, currently, you've giving and.

result = [a_tuple for a_tuple in L if a in a_tuple or b in a_tuple]

Upvotes: 0

Kyle Hannon
Kyle Hannon

Reputation: 2249

You can do this:

r = [i for i,x in enumerate(L) if any(y in x for y in (a,b))]

enumerate will give you both indices and values in your list comprehension, and the any statement will tell you if either a or b are in x, which is a sublist in L

Upvotes: 2

inetphantom
inetphantom

Reputation: 2617

Use any() to test the sublists:

if any(a in subl for subl in L):

This tests each subl but exits the generator expression loop early if a match is found.

This does not, however, return the specific sublist that matched. You could use next() with a generator expression to find the first match:

matched = next((subl for subl in L if a in subl), None)
if matched is not None:
    matched[1] += 1

where None is a default returned if the generator expression raises a StopIteration exception, or you can omit the default and use exception handling instead:

try:
    matched = next(subl for subl in L if a in subl)
    matched[1] += 1
except StopIteration:
    pass # no match found

Upvotes: 0

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