CODEWITHSUNDEEP
pythonregexpandas
GuyGuyGuy
GuyGuyGuy

Reputation: 75

Creating a year column in Pandas

I'm trying to create a year column with the year taken from the title column in my dataframe. This code works, but the column dtype is object. For example, in row 1 the year displays as [2013].

How can i do this, but change the column dtype to a float? 

year_list = []

for i in range(title_length):
    year = re.findall('\d{4}', wine['title'][i])
    year_list.append(year)

wine['year'] = year_list

Here is the head of my dataframe:

country   designation     points    province               title             year
Italy     Vulkà Bianco     87        Sicily     Nicosia 2013 Vulkà Bianco   [2013]

Upvotes: 2

Views: 216

Answers (2)

Jab
Jab

Reputation: 27485

re.findall returns a list of results. Use re.search

wine['year'] = [re.search('\d{4}', title)[0] for title in wine['title']]

better yet use pandas extract method.

wine['year'] = wine['title'].str.extract(r'\d{4}')

Definition

Series.str.extract(pat, flags=0, expand=True)

For each subject string in the Series, extract groups from the first match of regular expression pat.

Upvotes: 2

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626738

Instead of re.findall that returns a list of strings, you may use str.extract():

wine['year'] = wine['title'].str.extract(r'\b(\d{4})\b')

Or, in case you want to only match 1900-2000s years:

wine['year'] = wine['title'].str.extract(r'\b((?:19|20)\d{2})\b')

Note that the pattern in str.extract must contain at least 1 capturing group, its value will be used to populate the new column. The first match will only be considered, so you might have to precise the context later if need be.

I suggest using word boundaries \b around the \d{4} pattern to match 4-digit chunks as whole words and avoid partial matches in strings like 1234567890.

Upvotes: 2

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