how to slice a string in dataframe, start from left, based on different characters, such as ' /- . , I only want the first time this character shows up.
key name
1 McDonald's
2 CVS/PHARMACY
3 CVS/Store
4 WAL-MART
5 AMAZON.CO
expect result:
key name for_Group
1 McDonald's McDonald
2 CVS/PHARMACY CVS
3 CVS/Store CVS
4 WAL-MART WAL
5 AMAZON.CO AMAZON
I'm not sure if this need to use regular expression?
Option 1
str.split with expand=True
df['for_group'] = df.name.str.split(r"[\'\/\-\.]", expand=True)[0]
key name for_group
0 1 McDonald's McDonald
1 2 CVS/PHARMACY CVS
2 3 CVS/Store CVS
3 4 WAL-MART WAL
4 5 AMAZON.CO AMAZON
Option 2 (Best option)
str.extract (I personally prefer this one, it matches until it finds one of your desired stop characters)
df.name.str.extract(r'(.*?)[\'\/\-\.]', expand=False)
0 McDonald
1 CVS
2 CVS
3 WAL
4 AMAZON
The second option here is much faster:
df = pd.concat([df]*10000)
%timeit df.name.str.split(r"[\'\/\-\.]", expand=True)[0]
141 ms ± 1.8 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit df.name.str.extract(r'(.*)[\'\/\-\.]', expand=False)
72.6 ms ± 397 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
(.*)[\'\/\-\.] as the regex with extract. I updated it to (.*?)[\'\/\-\.] which should fix bad matches. I would recommend trying the current version in my answer and seeing if that works better for you!(.*)[\'\/\-\.] will match from the beginning of your string until the last delimiter, but you want lazy matching and by using (.*?) instead of (.*) it only matches until the first occurence of a delimiterMethod 1
You can use the below regular expression, which means a word character (a-z etc.) repeated one or more times. This returns an array and you can take the first element off it.
import re
df['for_group'] = df['name'].apply(lambda x: re.findall(r"[\w]+", x)[0])
A faster approach to regular expression would be to use .search() as pointed out by @user3483203
df['for_group'] = df['name'].apply(lambda x: re.search(r"[\w]+", x).group())
Method 2
Similarly, you can use:
df['for_group'] = df.name.str.split('\W+').apply(lambda x: x[0])
Output:
key name for_group
0 1 McDonald's McDonald
1 2 CVS/PHARMACY CVS
2 3 CVS/Store CVS
3 4 WAL-MART WAL
4 5 AMAZON.CO AMAZON
search as opposed to findall, so df['name'].apply(lambda x: re.search(r"[\w]+", x).group())
.findall() takes 73.8 ms ± 1.22 ms time while .search() takes 60.1 ms ± 552 µs. Thanks..split() and/or .findall() ?search and extract return after the first match is found, while split and findall will search the entire string. If the strings in the columns were longer search/extract would be even more desireable.