How to use two variables with one function in python?-CodePudding

I want a new column in a df that shows two options when executing a function.

I have the following lists:

lista = [A, B, C, D]
listb = [use to, manage to, when, did not]

If in a column called "Description" that contains texts in each cells is it found any word of "lista" return that value in a new column called "Effect" if this is not found, then search for values of "listb" and return that value in the same column "Effect" but with additional next 1 or 2 strings of that word, for example the word from listb "use to" with it next word fail in order I can get the complete setence "use to fail" in the column.

I have tried something like this:

def matcher(Description):
    for i in lista:
        if i in Description:
            return i
    return "Not found"

def matcher(Description):
    for j in listb:
        if j in Description:
            return j   1
    return "Not found"

df["Effect"] = df.apply(lambda i: matcher(i["Description"]), axis=1)
df["Effect"] = df.apply(lambda j: matcher(j["Description"]), axis=1)

CodePudding user response：

You can do both at once:

def matcher(Description):
    w = [i for i in lista if i in Description]
    w.extend( [i for i in listb if i in Description] )
    if not w:
        return "Not found"
    else:
        return ' '.join(w)

df["Effect"] = df.apply(lambda i: matcher(i["Description"]), axis=1)

CodePudding user response：

The code below should do what you want to achieve:

def matcher(sentence):
    match_list = [substr for substr in lista if substr in sentence.split(" ")]
    if match_list: # list with items evaluates to True
        return match_list[0]
    match_list = [substr for substr in listb if substr in sentence]
    if match_list:
        substr = match_list[0]
        return substr   " "   sentence.split(substr)[-1].strip().split(" ")[0]
    return "Not found"

df["Effect"] = df.Description.apply(matcher)