Partial string matching on large data set - python
So I have been working on this for a while and just haven't got any where and just not sure what to do.Fairly new to pandas and python.
Data set is actually 15,000 product names. All in different formats, some have multiple dashes up to 6, some hyphens, different lengths,The rows are product names with variants.
The code i'm using keeps returning only the first letter as oppose to the partial string when I use it on a large data set.
Works just fine on a small data set which I was using to test it.
I'm assuming this is happening because:
I haven't created a stop section when it matches a full partial string
because its trying to match up words as oppose to individual characters and stopping when it finds a difference.
What is the best way to overcome this on a large data set, what am I missing? or am I going to have to do this manual?
Original test data set
`1.star t-shirt-large-red
2.star t-shirt-large-blue
3.star t-shirt-small-red
4.beautiful rainbow skirt small
5.long maxwell logan jeans- light blue -32L-28W
6.long maxwell logan jeans- Dark blue -32L-28W`
Desired data set/output:
`COL1 COL2 COL3 COL4
1[star t-shirt] [large] [red] NONE
2[star t-shirt] [large] [blue] NONE
3[star t-shirt] [small] [red] NONE
4[beautiful rainbow skirt small] [small] NONE NONE
5[long maxwell logan jeans] [light blue] [32L] [28W]
6[long maxwell logan jeans] [Dark blue] [32L] [28W]`
Here is the code I was helped with in a previous question I asked:
`df['onkey'] = 1
df1 = pd.merge(df[['name','onkey']],df[['name','onkey']], on='onkey')
df1['list'] = df1.apply(lambda x:[x.name_x,x.name_y],axis=1)
from os.path import commonprefix
df1['COL1'] = df1['list'].apply(lambda x:commonprefix(x))
df1['COL1_num'] = df1['COL1'].apply(lambda x:len(x))
df1 = df1[(df1['COL1_num']!=0)]
df1 = df1.loc[df1.groupby('name_x')['COL1_num'].idxmin()]
df = df.rename(columns ={'name':'name_x'})
df = pd.merge(df,df1[['name_x','COL1']],on='name_x',how ='left')`
`df['len'] = df['COL1'].apply(lambda x: len(x))
df['other'] = df.apply(lambda x: x.name_x[x.len:],axis=1)
df['COL1'] = df['COL1'].apply(lambda x: x.strip())
df['COL1'] = df['COL1'].apply(lambda x: x[:-1] if x[-1]=='-' else x)
df['other'] = df['other'].apply(lambda x:x.split('-'))
df = df[['COL1','other']]
df = pd.concat([df['COL1'],df['other'].apply(pd.Series)],axis=1)`
` COL1 0 1 2
0 star t-shirt large red NaN
1 star t-shirt large blue NaN
2 star t-shirt small red NaN
3 beautiful rainbow skirt small NaN NaN
4 long maxwell logan jeans light blue 32L 28W
5 long maxwell logan jeans Dark blue 32L 28W`
***************update*****************
This is your input list of product,some have variants and some don't.
When searching for duplicates strings to determine what are the products with variants and products without variants;nothing comes up because they are all seen as unique values due to the variants being added on at the end of the string.
So what I would like to do is group the partial or similar strings together(the longest match), extract the longest matching string within the group and then put the differences into other columns.
If the product /string is unique just print into the column with the extracted longest string.
star t-shirt-large-red
star t-shirt-large-blue
star t-shirt-small-red
beautiful rainbow skirt small
long maxwell logan jeans- light blue -32L-28W
long maxwell logan jeans- Dark blue -32L-28W
Organic and natural candy - 3 Pack - Mint
Organic and natural candy - 3 Pack - Vanilla
Organic and natural candy - 3 Pack - Strawberry
Organic and natural candy - 3 Pack - Chocolate
Organic and natural candy - 3 Pack - Banana
Organic and natural candy - 3 Pack - Cola
Organic and natural candy - 12 Pack Assorted
Morgan T-shirt Company - Small/Medium-Blue
Morgan T-shirt Company - Medium/Large-Blue
Morgan T-shirt Company - Medium/Large-red
Morgan T-shirt Company - Small/Medium-Red
Morgan T-shirt Company - Small/Medium-Green
Morgan T-shirt Company - Medium/Large-Green
Nelly dress leopard small
col1 col2 col3 col4
star t-shirt large red
star t-shirt large blue
star t-shirt small red
beautiful rainbow skirt small
Long maxwell logan jeans light blue 32L 28W
Long maxwell logan jeans Dark blue 32L 28W
Organic and natural candy 3 Pack Mint
Organic and natural candy 3 Pack Vanilla
Organic and natural candy 3 Pack Strawberry
Organic and natural candy 3 Pack Chocolate
Organic and natural candy 3 Pack Banana
Organic and natural candy 3 Pack Cola
Organic and natural candy 12 Pack Assorted
Morgan T-shirt Company Small/Medium Blue
Morgan T-shirt Company Medium/Large Blue
Morgan T-shirt Company Medium/Large Red
Morgan T-shirt Company Small/Medium Red
Morgan T-shirt Company Small/Medium Green
Morgan T-shirt Company Medium/Large Green
Nelly dress Leopard Small
Bijoux
Princess PJ-set
Lemon tank top Yellow Medium
Constructing a DataFrame df as follows:
df = pd.DataFrame()
df = df.append(['1.star t-shirt-large-red'])
df = df.append(['2.star t-shirt-large-blue'])
df = df.append(['4.beautiful rainbow skirt small'])
df = df.append(['5.long maxwell logan jeans- light blue -32L-28W'])
df = df.append(['6.long maxwell logan jeans- Dark blue -32L-28W'])
df.columns = ['Product']
The following code
(a) strips any whitespace,
(b) splits by the period ('.') and grabs what follows,
(c) replaces 't-shirt' with 'tshirt' because of further operations (change this back if you want after the operation)
(d) splits again by '-' and expands to give your dataframe.
df['Product'].str.strip().str.split('.').str.get(1).str.replace('t-shirt', 'tshirt').str.split('-', expand = True)
Output:
0 1 2 3
0 star tshirt large red None
0 star tshirt large blue None
0 beautiful rainbow skirt small None None None
0 long maxwell logan jeans light blue 32L 28W
0 long maxwell logan jeans Dark blue 32L 28W
Given the inconsistency in nomenclature for your product, there will be edge-cases that are missed (ex : beautiful rainbow skirt small). You may have to fish them out again.
A solution which is quite simple to understand, debug and flexibly extend is the following:
Consider that your initial product names are held in a list called strings.
Then the solution is the following line:
mydf = pd.concat([pd.DataFrame([make_row(row, 4)], columns=['COL1', 'COL2', 'COL3', 'COL4']) for row in strings], ignore_index=True)
where we have defined the parsing function make_row to be:
def make_row(string, num_cols):
cols = [item.strip() for item in string[2:].split('-')] # ignore numbering, split on hyphen and strip whitespace
if len(cols) < num_cols:
cols += [np.nan]*(num_cols - len(cols)) # fill with NaN missing values
return cols
The first line defining cols could also be simply cols = string.split('-'), in which case you could do the formatting afterwards with:
mydf.applymap(lambda x: x if pd.isnull(x) else str.strip(x))
Now in your case, I see that there is a hyphen in some of your product names, in which case you might want to 'sanitize' them in advance (or inside make_row, as you wish), with something like:
strings = [item.replace('t-shirt', 'tshirt') for item in strings]
Example input:
strings = ['1.one-two-three', '2. one-two', '3.one-two-three-four', '4.one - two -three -four ']
Output:
COL1 COL2 COL3 COL4
0 one two three NaN
1 one two NaN NaN
2 one two three four
3 one two three four
Output for question's data (after correcting typo for item 4):
COL1 COL2 COL3 COL4
0 star tshirt large red NaN
1 star tshirt large blue NaN
2 star tshirt small red NaN
3 beautiful rainbow skirt small NaN NaN
4 long maxwell logan jeans light blue 32L 28W
5 long maxwell logan jeans Dark blue 32L 28W
Edit:
If you additionally want to "group" the items together, then you can:
a) Use sort_values (pandas doc) on the column COL1 after you get a dataframe as described above to simply display the rows corresponding to the same product the one after the other, or
b) use group_by to actually get a grouped dataframe like this:
grouped_df = mydf.groupby("COL1")
This will allow you to get each group like this:
grouped_df.get_group("star tshirt")
Producing following output:
COL1 COL2 COL3 COL4
0 star tshirt large red NaN
1 star tshirt large blue NaN
2 star tshirt small red NaN
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