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Need help running a full match between two large dataframes based off the full business name compared to the description of a newsfeed

I have two dataframes:
One with a single column of business names that I call 'bus_names_2' with a column name of 'BUSINESS_NAME'
One with an array of records and fields that was pulled from a RSS feed that I call 'df_newsfeed'. The import field is 'Description_2' field which represents the RSS feeds contents after scrubbing stopwords and symbols. This was also conducted on the 'bus_names_2' dataframe as well.
I am trying to look through each record in the 'df_newsfeed's 'Description_2' field to see if any array of words contains a business name from the 'bus_names_2' dataframe. This is easily done using the following:
def IdentityResolution_demo(bus_names, df, col='Description_2', upper=True):
n_rows = df.shape[0]
description_col = df.columns.get_loc(col)
df['Company'] = ''
company_col = df.columns.get_loc('Company')
if upper:
df.loc[:,col] = df.loc[:,col].str.upper()
for ind in range(n_rows):
businesses = []
description = df.iloc[ind,description_col]
for bus_name in bus_names:
if bus_name in description:
businesses.append(bus_name)
if len(businesses) > 0:
company = '|'.join(businesses)
df.iloc[ind,company_col] = company
df = df[['Source', 'RSS', 'Company', 'Title', 'PublishedDate', 'Description', 'Link']].drop_duplicates()
return df
bus_names_3 = list(set(bus_names_2['BUSINESS_NAME'].tolist()))
test = IdentityResolution_demo(bus_names_3, df_newsfeed.iloc[:10])
test[test['Company']!='']
This issue with this asides from the length of time it takes is that it is bringing back everything in a contains manner. I only want full word matches. Meaning if I have a company in my 'bus_names_2' dataframe called 'Bank of A' that it only brings back that name into the company category if the full word of 'Bank of A' exist in the 'Description_2' column of the 'df_newsfeed' dataframe and not when 'Bank of America' shows up.
Essentially, I need something like this ingrained in my function to produce the proper output for the 'Company' column but I don't know how to implement it. The below code gets the point accross.
Description_2 = 'GUARDFORCE AI CO LIMITED AI GFAIW RIVERSOFT INC PEAKWORK COMPANY GFAIS CONCIERGE GUARDFORCE AI RIVERSOFT ROBOT TRAVEL AGENCY'
bus_name_2 = ['GUARDFORCE AI CO']
for i in bus_name_2:
bus_name = re.compile(fr'\b{i}\b')
print(f"{i if bus_name.match(Description_2) else ''}")
This would produce an output of 'GUARDFORCE AI CO' but if I change the bus_name_2 to:
bus_name_2 = ['GUARDFORCE AI C']
It would produce a null output.
This function is written in the way it is because comparing two dataframes turned into a very long query and so optimization required a non-dataframe format.

Pandas map two dataframes using regex

I've two dataframes, one with text information and another with regex and patterns, what I need to do is to map a column from the second dataframe using regex
edit: What I need to do is to apply each regex on all df['text'] rows, and if there is a match, add the Pattern into a new column
Sample data
text_dict = {'text':['customer and increased repair and remodel activity as well as from other sales',
'sales for the overseas customers',
'marketing approach is driving strong play from top tier customers',
'employees in India have been the continuance of remote work will impact productivity',
'sales due to higher customer']}
regex_dict = {'Pattern':['Sales + customer', 'Marketing + customer', 'Employee * Productivity'],
'regex': ['(?:sales\\w*)(?:[^,.?])*(?:customer\\w*)|(?:customer\\w*)(?:[^,.?])*(?:sales\\w*)',
'(?:marketing\\w*)(?:[^,.?])*(?:customer\\w*)|(?:customer\\w*)(?:[^,.?])*(?:marketing\\w*)',
'(?:employee\\w*)(?:[^\n])*(?:productivity\\w*)|(?:productivity\\w*)(?:[^\n])*(?:employee\\w*)']}
df
text
0 customer and increased repair and remodel acti...
1 sales for the overseas customers
2 marketing approach is driving strong play from...
3 employees in India have been the continuance o...
4 sales due to higher customer
regex
Pattern regex
0 Sales + customer (?:sales\w*)(?:[^,.?])*(?:customer\w*)|(?:cust...
1 Marketing + customer (?:marketing\w*)(?:[^,.?])*(?:customer\w*)|(?:...
2 Employee * Productivity (?:employee\w*)(?:[^\n])*(?:productivity\w*)|(...
Desired output
text Pattern
0 customer and increased repair and remodel acti... Sales + customer
1 sales for the overseas customers Sales + customer
2 marketing approach is driving strong play from... Marketing + customer
3 employees in India have been the continuance o... Employee * Productivity
4 sales due to higher customer Sales + customer
tried the following, created a function that returns the Pattern in case there is a match, then I iterate over all the columns in the regex dataframe
def finding_keywords(regex, match, keyword):
if re.search(regex, match):
return keyword
else:
pass
for index, row in regex.iterrows():
df['Pattern'] = df['text'].apply(lambda x: finding_keywords(regex['Regex'][index], x, regex['Pattern'][index]))
the problem with this is that in every iteration, it erases the previous mappings, as you can see below. As I'm foo foo was the last iteration, is the only one remaining with a pattern
text Pattern
0 foo None
1 bar None
2 foo foo I'm foo foo
3 foo bar None
4 bar bar None
One solution could be to run the iteration over regex dataframe, and then iterate over df, this way I avoid loosing information, but I'm looking for a fastest solution

You can loop through the unique values of the regex dataframe and apply to the text of the df frame and return the pattern in a new regex column. Then, merge in the Pattern column and drop the regex column.
The key to my approach was to first create the column as NaN and then fillna with each iteration so the columns didn't get overwritten.
import re
import numpy as np
srs = regex['regex'].unique()
df['regex'] = np.nan
for reg in srs:
df['regex'] = df['regex'].fillna(df['text'].apply(lambda x: reg
if re.search(reg, x) else np.NaN))
df = pd.merge(df, regex, how='left', on='regex').drop('regex', axis=1)
df
Out[1]:
text Pattern
0 customer and increased repair and remodel acti... Sales + customer
1 sales for the overseas customers Sales + customer
2 marketing approach is driving strong play from... Marketing + customer
3 employees in India have been the continuance o... Employee * Productivity
4 sales due to higher customer Sales + customer

Pandas: Replacing column values with ones as retrieved from other dataframe

I am stumbled upon a trivial problem in pandas. I have two dataframes. The first one, df_1 is as follows
vendor_name date company_name state
PERTH is june 2019 Abc enterprise Kentucky
Megan Ent 25-april-2019 Xyz Fincorp Texas
The second one df_2 contains the correct values for each column in df_1.
df_2
Field wrong value correct value
vendor_name PERTH Perth Enterprise
date is 15 ## this means that is should be read as 15
company_name Abc enterprise ABC International Enterprise Inc.
In order to replace the values with correct ones in df_1 (except date field) I am using pandas.loc method. Below is the code snippet
vend = df_1['vendor_Name'].tolist()
comp = df_1['company_name'].tolist()
state = df_1['state'].tolist()
for i in vend:
if df_2['wrong value'].str.contains(i):
crct = df_2.loc[df_2['wrong value'] == i,'correct value'].tolist()
Similarly, for company and state I have followed the above way.
However, the crct is returning a blank series. Ideally it should return
['Perth Enterprise','Abc International Enterprise Inc']
The next step would be to replace the respective field values by the above list.
With the above, I have three questions:
Why the above code is generating a blank list? What I am missing here?
How can I replace the respective fields using df_1.replace method?
What should be a correct approach to replace the portion of date in df_1 by the correct one in df_2?

Edit: when data has looping replacement(i.e overlaping keys and values), replacement on whole dataframe will fail. In this case, doing it column by column and concat them together. Finally, use join to adding any missing columns from df1:
df_replace = pd.concat([df1[k].replace(val, regex=True) for k, val in d.items()], axis=1).join(df1.state)
Original:
I tried your code in my interactive and it gives error ValueError: The truth value of a Series is ambiguous on df_2['wrong value'].str.contains(i).
assume you have multiple vendor names, so the simple way is construct a dictionary from groupby of df2 and use it with df.replace on df1.
d = {k: gp.set_index('wrong value')['correct value'].to_dict()
for k, gp in df2.groupby('Field')}
Out[64]:
{'company_name': {'Abc enterprise': 'ABC International Enterprise Inc. '},
'date': {'is': '15'},
'vendor_name': {'PERTH': 'Perth Enterprise'}}
df_replace = df1.replace(d, regex=True)
print(df_replace)
In [68]:
vendor_name date company_name \
0 Perth Enterprise 15 june 2019 ABC International Enterprise Inc.
1 Megan Ent 25-april-2019 Xyz Fincorp
state
0 Kentucky
1 Texas
Note: your sample df2 has only value for vendor PERTH, so it only replace first row. When you have all vendor_names in df2, it will replace them all in df1.

A simple way to do that is to iterate over the first dataframe and then replace the wrong values :
Result = pd.DataFrame()
for i in range(len(df1)):
vendor_name = df1.iloc[i]['vendor_name']
date = df1.iloc[i]['date']
company_name = df1.iloc[i]['company_name']
if vendor_name in df2['wrong value'].values:
vendor_name = df2.loc[df2['wrong value'] == vendor_name]['correct value'].values[0]
if company_name in df2['wrong value'].values:
company_name = df2.loc[df2['wrong value'] == company_name]['correct value'].values[0]
new_row = {'vendor_name':[vendor_name],'date':[date],'company_name':[company_name]}
new_row = pd.DataFrame(new_row,columns=['vendor_name','date','company_name'])
Result = Result.append(new_row,ignore_index=True)
Result :

Define the following replace function:
def repl(row):
fld = row.Field
v1 = row['wrong value']
v2 = row['correct value']
updInd = df_1[df_1[fld].str.contains(v1)].index
df_1.loc[updInd, fld] = df_1.loc[updInd, fld]\
.str.replace(re.escape(v1), v2)
Then call it for each row in df_2:
for _, row in df_2.iterrows():
repl(row)
Note that str.replace alone does not require to import re (Pandas
imports it under the hood).
But in the above function re.escape is called explicitely, from our code,
hence import re is required.

strip data frame cell then create columns

i'm trying to take the info from dataframe and break it out into columns with the following header names. the info is all crammed into 1 cell.
new to python, so be gentle.
thanks for the help
my code:
r=requests.get('https://nclbgc.org/search/licenseDetails?licenseNumber=80479')
page_data = soup(r.text, 'html.parser')
company_info = [' '.join(' '.join(info.get_text(", ", strip=True).split()) for info in page_data.find_all('tr'))]
df = pd.DataFrame(company_info, columns = ['ic_number, status, renewal_date, company_name, address, county, telephon, limitation, residential_qualifiers'])
print(df)
the result i get:
['License Number, 80479 Status, Valid Renewal Date, n/a Name, DLR Construction, LLC Address, 3217 Vagabond Dr Monroe, N
C 28110 County, Union Telephone, (980) 245-0867 Limitation, Limited Classifications, Residential Qualifiers, Arteaga, Vi
cky Rodriguez']

You can use read_html with some post processing:
url = 'https://nclbgc.org/search/licenseDetails?licenseNumber=80479'
#select first table form list of tables, remove only NaNs rows
df = pd.read_html(url)[0].dropna(how='all')
#forward fill NaNs in first column
df[0] = df[0].ffill()
#merge values in second column
df = df.groupby(0)[1].apply(lambda x: ' '.join(x.dropna())).to_frame().rename_axis(None).T
print (df)
Address Classifications County License Number \
1 3217 Vagabond Dr Monroe, NC 28110 Residential Union 80479
Limitation Name Qualifiers Renewal Date \
1 Limited DLR Construction, LLC Arteaga, Vicky Rodriguez
Status Telephone
1 Valid (980) 245-0867

Replace the df line like below:
df = pd.DataFrame(company_info, columns = ['ic_number', 'status', 'renewal_date', 'company_name', 'address', 'county', 'telephon', 'limitation', 'residential_qualifiers'])
Each column mentioned under columns should be within quotes. Else it is considered as one single column.

Pandas - Matching reference number to find earliest date

I'm hoping to pick your brains on optimization. I am still learning more and more about python and using it for my day to day operation analyst position. One of the tasks I have is sorting through approx 60k unique record identifiers, and searching through another dataframe that has approx 120k records of interactions, the employee who authored the interaction and the time it happened.
For Reference, the two dataframes at this point look like:
main_data = Unique Identifier Only
nok_data = Authored By Name, Unique Identifer(known as Case File Identifier), Note Text, Created On.
My set up currently runs it at approximately sorting through and matching my data at 2500 rows per minute, so approximately 25-30 minutes or so for a run. What I am curious is are there any steps I performed that are:
Redundant and inefficient overall slowing my process
A poor use of syntax to work around my lack of knowledge.
Below is my code:
nok_data = pd.read_csv("raw nok data.csv") #Data set from warehouse
main_data = pd.read_csv("exampledata.csv") #Data set taken from iTx ids from referral view
row_count = 0
error_count = 0
print(nok_data.columns.values.tolist())
print(main_data.columns.values.tolist()) #Commented out, used to grab header titles if needed.
data_length = len(main_data) #used for counting how many records left.
earliest_nok = {}
nok_data["Created On"] = pd.to_datetime(nok_data["Created On"]) #convert all dates to datetime at beginning.
for row in main_data["iTx Case ID"]:
list_data = []
nok = nok_data["Case File Identifier"] == row
matching_dates = nok_data[["Created On", "Authored By Name"]][nok == True] #takes created on date only if nok shows row was true
if len(matching_dates) > 0:
try:
min_dates = matching_dates.min(axis=0)
earliest_nok[row] = [min_dates[0], min_dates[1]]
except ValueError:
error_count += 1
earliest_nok[row] = None
row_count += 1
print("{} out of {} records").format(row_count, data_length)
with open('finaloutput.csv','wb') as csv_file:
writer = csv.writer(csv_file)
for key, value in earliest_nok.items():
writer.writerow([key, value])
Looking for any advice or expertise from those performing code like this much longer then I have. I appreciate all of you who even just took the time to read this. Happy Tuesday,
Andy M.
**** EDIT REQUESTED TO SHOW DATA
Sorry for my novice move there not including any data type.
main_data example
ITX Case ID
2017-023597
2017-023594
2017-023592
2017-023590
nok_data aka "raw nok data.csv"
Authored By: Case File Identifier: Note Text: Authored on
John Doe 2017-023594 Random Text 4/1/2017 13:24:35
John Doe 2017-023594 Random Text 4/1/2017 13:11:20
Jane Doe 2017-023590 Random Text 4/3/2017 09:32:00
Jane Doe 2017-023590 Random Text 4/3/2017 07:43:23
Jane Doe 2017-023590 Random Text 4/3/2017 7:41:00
John Doe 2017-023592 Random Text 4/5/2017 23:32:35
John Doe 2017-023592 Random Text 4/6/2017 00:00:35

It looks like you want to group on the Case File Identifier and get the minimum date and corresponding author.
# Sort the data by `Case File Identifier:` and `Authored on` date
# so that you can easily get the author corresponding to the min date using `first`.
nok_data.sort_values(['Case File Identifier:', 'Authored on'], inplace=True)
df = (
nok_data[nok_data['Case File Identifier:'].isin(main_data['ITX Case ID'])]
.groupby('Case File Identifier:')['Authored on', 'Authored By:'].first()
)
d = {k: [v['Authored on'], v['Authored By:']] for k, v in df.to_dict('index').iteritems()}
>>> d
{'2017-023590': ['4/3/17 7:41', 'Jane Doe'],
'2017-023592': ['4/5/17 23:32', 'John Doe'],
'2017-023594': ['4/1/17 13:11', 'John Doe']}
>>> df
Authored on Authored By:
Case File Identifier:
2017-023590 4/3/17 7:41 Jane Doe
2017-023592 4/5/17 23:32 John Doe
2017-023594 4/1/17 13:11 John Doe
It is probably easier to use df.to_csv(...).
The items from main_data['ITX Case ID'] where there is no matching record have been ignored but could be included if required.