Searching many strings for many dictionary keys, quickly

Question

I have a unique question, and I am primarily hoping to identify ways to speed up this code a little. I have a set of strings stored in a dataframe, each of which has several names in it and I know the number of names before this step, like so:

print df

description                      num_people        people    
'Harry ran with sally'                2              []         
'Joe was swinging with sally'         2              []
'Lola Dances alone'                   1              []

I am using a dictionary with the keys that I am looking to find in description, like so:

my_dict={'Harry':'1283','Joe':'1828','Sally':'1298', 'Cupid':'1982'}

and then using iterrows to search each string for matches like so:

for index, row in df.iterrows():
    row.people=[key for key in my_dict if re.findall(key,row.desciption)]

and when run it ends up with

print df

 description                      num_people        people    
'Harry ran with sally'                2              ['Harry','Sally']         
'Joe was swinging with sally'         2              ['Joe','Sally']
'Lola Dances alone'                   1              ['Lola']

The problem that I see, is that this code is still fairly slow to get the job done, and I have a large number of descriptions and over 1000 keys. Is there a faster way of performing this operation, like maybe using the number of people found?

jezrael · Accepted Answer · 2016-09-28 08:17:24Z

Faster solution:

#strip ' in start and end of text, create lists from words
splited = df.description.str.strip("'").str.split()
#filtering
df['people'] = splited.apply(lambda x: [i for i in x if i in my_dict.keys()])
print (df)
                     description  num_people          people
0         'Harry ran with Sally'           2  [Harry, Sally]
1  'Joe was swinging with Sally'           2    [Joe, Sally]
2            'Lola Dances alone'           1          [Lola]

Timings:

#[30000 rows x 3 columns]
In [198]: %timeit (orig(my_dict, df))
1 loop, best of 3: 3.63 s per loop

In [199]: %timeit (new(my_dict, df1))
10 loops, best of 3: 78.2 ms per loop

df['people'] = [[],[],[]]
df = pd.concat([df]*10000).reset_index(drop=True)
df1 = df.copy()

my_dict={'Harry':'1283','Joe':'1828','Sally':'1298', 'Lola':'1982'}

def orig(my_dict, df):
    for index, row in df.iterrows():
        df.at[index, 'people']=[key for key in my_dict if re.findall(key,row.description)]
    return (df)


def new(my_dict, df):
    df.description = df.description.str.strip("'")
    splited = df.description.str.split()
    df.people = splited.apply(lambda x: [i for i in x if i in my_dict.keys()])
    return (df)


print (orig(my_dict, df))
print (new(my_dict, df1))

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