Assigning True/False if a token is present in a data-frame

try

df["trumpMention"] = df["keywords"].apply(lambda x: "Trump, Donald J" in x)

How about applying a function that checks set membership?

df['trumpMention'] = df['keywords'].apply(lambda x: 'Trump, Donald J' in set(x))

Output:

  articleID                                           keywords  trumpMention
0  58b61d1d                    [Second Avenue (Manhattan, NY)]         False
1  58b6393b                                [Crossword Puzzles]         False
2  58b6556e  [Workplace Hazards and Violations, Trump, Dona...          True
3  58b657fa         [Trump, Donald J, Speeches and Statements]          True

As to your attempts:

np.where(any(df['keywords']) == 'Trump, Donald J', True, False) 

wouldn't work because any(df['keywords']) would always evaluate True which isn't equal to 'Trump, Donald J', so the above will always return array(False).

df['keywords'].apply(lambda x: any(token == 'Trump, Donald J') for token in x) 

doesn't work because it raises TypeError since there is no comprehension here.

df['keywords'].apply(lambda x: ([ True for token in x if any(token in lst)]))  

doesn't work because token in lst is a boolean value, so

any(token in lst)

is nonsensical.

Use a vectorized approach, it will be faster than using apply.

df.keywords.astype(str).str.contains("Trump, Donald J")

Try my way. I create a list before adding it to dataframe.

def mentioned_Trump(s, lst):
    if s in lst:
        return True
    else:
        return False
s = [[1,['Second Avenue (Manhattan, NY)']],[2,['Crossword Puzzles']],
    [3, ['Workplace Hazards and Violations', 'Trump, Donald J']],
    [4, ['Trump, Donald J', 'Speeches and Statements']]]

import pandas as pd
df = pd.DataFrame(s)
df.columns =['ID','keywords']

s = list( df['keywords'])
s1 = [mentioned_Trump('Trump, Donald J',x) for x in s]

df['trumpMention']= s1 
print(df)