CODEWITHSUNDEEP
pythonpython-3.xlist
Vinay Prakash
Vinay Prakash

Reputation: 35

How to search if a number from a list is present in another list?

I have a list that contains some numbers. I have to search those numbers one by one to check if they are present in 2nd list. But I have to check number from 1st list completely i.e if full match is not there, then remove the last digit from the number and search again till match is found. I have to keep doing this for all digits of each number of list 1.

example:

z = ['48761', '3876', '481']

p = ['3876112','8935']

I pick 3876112... seach for it in list z. if no match, I search 387611, if no match... I seach for 38761 and so on... Any time in the process if a match is found, I have to return match found and then start doing same for 8935.

Note: If there is no match at all, I need to return "No match" and that is only Once for that number.

z = ['48761', '3876', '481']

p = ['3876112','8935']

for m in range(0,len(p)):
    x = p[m]
    for i in range(0,len(x)):
        y = x
        for words in z:
            if y == words:
                print("Found")      
            x = x[:-1]

    print("Not found")

Result:

Found
Not found
Not found

Also, When I added in between print lines, I noticed with above code, it is running loop one extra time before printing Found.

Upvotes: 0

Views: 263

Answers (4)

Tomerikoo
Tomerikoo

Reputation: 19414

Since you're working with strings and not actual numbers, you could use str.startswith():

z = ['48761', '3876', '481']

p = ['3876112','8935']

res = []
for target in p:
    for num in z:
        if target.startswith(num):
            res.append("Found")
            break   
    else:
        res.append("Not found")

Or simply:

res= ["Found" if any(target.startswith(num) for num in z) else "Not Found" for target in p]

And now, assuming your dataframe is df you can simply do:

df["res"] = res

Upvotes: 3

Nikos Oikou
Nikos Oikou

Reputation: 241

You can try something like this:

p = ['3876112','8935']
z = ['48761', '3876', '481']

z = set(z)  # Cast to set for efficient membership search

for elem in p:
    found = False

    for i in reversed(range(1, len(elem))):

        if elem[:i] in z:
            found = True
            print('Found')
            break

    if not found:
        print('Not found')

The output is:

Found
Not found

thus one output line for each input element of the list p.

Upvotes: 1

Andrej Kesely
Andrej Kesely

Reputation: 195418

Using str.startswith() and any() to short-circuit early: 

z = ['48761', '3876', '481']
p = ['3876112','8935']

for num in p:
    if any(num.startswith(n) for n in z):
        print('Found')
    else:
        print('Not Found')

Prints:

Found
Not Found

Upvotes: 1

yaho cho
yaho cho

Reputation: 1779

Use slice of string.

z = ['48761', '3876', '481']
p = ['3876112','8935']

for p_ in p:
    length = len(p_)
    found = False
    save = None
    for i in range(length):
        if p_[:length-i] in z: 
            found = True
            save = p_[:length-i]
    if found: print('found', save)
    else: print('not found')

The result is:

found 3876
not found

Upvotes: 1

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