I have gotten a very strange data. I have dictionary with keys and values where I want to use this dictionary to search if these keywords are ONLY starting and/or end of the text not middle of the sentence. I tried to create simple data frame below to show the problem case and python codes that I have tried so far. How do I get it go search for only starting or ending of the sentence? This one searches whole text sub-strings.
Code:
d = {'apple corp':'Company','app':'Application'} #dictionary
l1 = [1, 2, 3,4]
l2 = [
"The word Apple is commonly confused with Apple Corp which is a business",
"Apple Corp is a business they make computers",
"Apple Corp also writes App",
"The Apple Corp also writes App"
]
df = pd.DataFrame({'id':l1,'text':l2})
df['text'] = df['text'].str.lower()
df
Original Dataframe:
id text
1 The word Apple is commonly confused with Apple Corp which is a business
2 Apple Corp is a business they make computers
3 Apple Corp also writes App
4 The Apple Corp also writes App
Code Tried out:
def matcher(k):
x = (i for i in d if i in k)
# i.startswith(k) getting error
return ';'.join(map(d.get, x))
df['text_value'] = df['text'].map(matcher)
df
Error:
TypeError: 'in <string>' requires string as left operand, not bool
when I use this x = (i for i in d if i.startswith(k) in k)
Empty values if i tried this x = (i for i in d if i.startswith(k) == True in k)
TypeError: sequence item 0: expected str instance, NoneType found
when i use this x = (i.startswith(k) for i in d if i in k)
Results from Code above ... Create new field 'text_value':
id text text_value
1 The word Apple is commonly confused with Apple Corp which is a business Company;Application
2 Apple Corp is a business they make computers Company;Application
3 Apple Corp also writes App Company;Application
4 The Apple Corp also writes App Company;Application
Trying to get an FINAL output like this:
id text text_value
1 The word Apple is commonly confused with Apple Corp which is a business NaN
2 Apple Corp is a business they make computers Company
3 Apple Corp also writes App Company;Application
4 The Apple Corp also writes App Application
You need a matcher function which can accept flag and then call that twice to get the results for startswith and endswith.
def matcher(s, flag="start"):
if flag=="start":
for i in d:
if s.startswith(i):
return d[i]
else:
for i in d:
if s.endswith(i):
return d[i]
return None
df['st'] = df['text'].apply(matcher)
df['ed'] = df['text'].apply(matcher, flag="end")
df['text_value'] = df[['st', 'ed']].apply(lambda x: ';'.join(x.dropna()),1)
df = df[['id','text', 'text_value']]
The text_value column looks like:
0
1 Company
2 Company;Application
3 Application
Name: text_value, dtype: object
joined = "|".join(d.keys())
pat = '(?i)^(?:the\\s*)?(' + joined + ')\\b.*?|.*\\b(' + joined + ')$'+'|.*'
get = lambda x: d.get(x.group(1),"") + (';' +d.get(x.group(2),"") if x.group(2) else '')
df.text.str.replace(pat,get)
0
1 Company
2 Company;Application
3 Company;Application
Name: text, dtype: object
Related
I am having troubles with updating a relationship property:
my goal is to map a dataset like the following into a Neo4j graph:
PersonName IllnessType
0 A 1
1 A 2
2 A 3
3 B 1
4 B 2
5 B 1
I basically cycle over the lines of this dataset, creating a Node for each Person and each Illness found on the line, and merging to avoid duplicates:
from py2neo import *
graph = Graph()
person_node= Node("Person", **kwargs)
graph.merge(person_node, "Person", "Name")
illness_node = Node("Illness", **kwargs)
graph.merge(illness_node, "Illness", "IllnessType")
edge = Relationship.type("SUFFERS_FROM")
rel = sfEdge(person_node, illness_node)
self.graph.merge(rel)
What I like to add now, is to add a weight on the "SUFFERS_FROM" edge that count how many times a person has suffered from a certain illness. What I tried to do was:
rm = RelationshipMatcher()
edge_to_increment = rm.match(nodes=(None, patNode), r_type=None).first()
if edge_to_increment is None:
edge_to_increment = edge(person_node, illness_node)
edge_to_increment["COUNT"]=1
self.graph.merge(edge_to_increment)
else:
edge_to_increment["COUNT"] += 1
c = e2r["COUNT"]
But then when I visualize the result, all edges have weight 1 even though the edge B-->1 should have weight 2.
Thanks in advance
So I have these strings that I split by spaces (' ') and I just rolled them into a single list I called 'keyLabelRun'
so it looks like this:
keyLabelRun[0-12]:
0 OS=Dengue
1 virus
2 3
3 PE=4
4 SV=1
5 Split=0
6
7 OS=Bacillus
8 subtilis
9 XF-1
10 GN=opuBA
11 PE=4
12 SV=1
I only want the elements that include and are after "OS=", anything else, whether it be "SV=" or "PE=" etc. I want to skip over those elements until I get to the next "OS="
The number of elements to the next "OS=" is arbitrary so that's where I'm having the problem.
This is what I'm currently trying:
OSarr = []
for i in range(len(keyLabelrun)):
if keyLabelrun[i].count('OS='):
OSarr.append(keyLabelrun[i])
if keyLabelrun[i+1].count('=') != 1:
continue
But the elements where "OS=" is not included is what is tripping me up I think.
Also at the end I'm going to join them all back together in their own elements but I feel like I will be able to handle that after this.
In my attempt, I am trying to append all elements I'm looking for in order to an new list 'OSarr'
If anyone can lend a hand, it would be much appreciated.
Thank you.
These list of strings came from a dataset that is a text file in the form:
>tr|W0FSK4|W0FSK4_9FLAV Genome polyprotein (Fragment) OS=Dengue virus 3 PE=4 SV=1 Split=0
MNNQRKKTGKPSINMLKRVRNRVSTGSQLAKRFSKGLLNGQGPMKLVMAFIAFLRFLAIPPTAGVLARWGTFKKSGAIKVLKGFKKEISNMLSIINKRKKTSLCLMMILPAALAFHLTSRDGEPRMIVGKNERGKSLLFKTASGINMCTLIAMDLGEMCDDTVTYKCPHITEVEPEDIDCWCNLTSTWVTYGTCNQAGEHRRDKRSVALAPHVGMGLDTRTQTWMSAEGAWRQVEKVETWALRHPGFTILALFLAHYIGTSLTQKVVIFILLMLVTPSMTMRCVGVGNRDFVEGLSGATWVDVVLEHGGCVTTMAKNKPTLDIELQKTEATQLATLRKLCIEGKITNITTDSRCPTQGEATLPEEQDQNYVCKHTYVDRGWGNGCGLFGKGSLVTCAKFQCLEPIEGKVVQYENLKYTVIITVHTGDQHQVGNETQGVTAEITPQASTTEAILPEYGTLGLECSPRTGLDFNEMILLTMKNKAWMVHRQWFFDLPLPWTSGATTETPTWNRKELLVTFKNAHAKKQEVVVLGSQEGAMHTALTGATEIQNSGGTSIFAGHLKCRLKMDKLELKGMSYAMCTNTFVLKKEVSETQHGTILIKVEYKGEDVPCKIPFSTEDGQGKAHNGRLITANPVVTKKEEPVNIEAEPPFGESNIVIGIGDNALKINWYKKGSSIGKMFEATARGARRMAILGDTAWDFGSVGGVLNSLGKMVHQIFGSAYTALFSGVSWVMKIGIGVLLTWIGLNSKNTSMSFSCIAIGIITLYLGAVVQADMGCVINWKGKELKCGSGIFVTNEVHTWTEQYKFQADSPKRLATAIAGAWENGVCGIRSTTRMENLLWKQIANELNYILWENNIKLTVVVGDIIGVLEQGKRTLTPQPMELKYSWKTWGKAKIVTAETQNSSFIIDGPNTPECPSVSRAWNVWEVEDYGFGVFTTNIWLKLREVYTQLCDHRLMSAAVKDERAVHADMGYWIESQKNGSWKLEKASLIEVKTCTWPKSHTLWSNGVLESDMIIPKSLAGPISQHNHRPGYHTQTAGPWHLGKLELDFNYCEGTTVVITENCGTRGPSLRTTTVSGKLIHEWCCRSCTLPPLRYMGEDGCWYGMEIRPISEKEENMVKSLVSAGSGKVDNFTMGVLCLAILFEEVMRGKFGKKHMIAGVFFTFVLLLSGQITWRDMAHTLIMIGSNASDRMGMGVTYLALIATFKIQPFLALGFFLRKLTSRENLLLGVGLAMATTLQLPEDIEQMANGIALGLMALKLITQFETYQLWTALISLTCSNTIFTLTVAWRTATLILAGVSLLPVCQSSSMRKTDWLPMAVAAMGVPPLPLFIFGLKDTLKRRSWPLNEGVMAVGLVSILASSLLRNDVPMAGPLVAGGLLIACYVITGTSADLTVEKAADITWEEEAEQTGVSHNLMITVDDDGTMRIKDDETENILTVLLKTALLIVSGIFPYSIPATLLVWHTWQKQTQRSGVLWDVPSPPETQKAELEEGVYRIKQQGIFGKTQVGVGVQKEGVFHTMWHVTRGAVLTYNGKRLEPNWASVKKDLISYGGGWRLSAQWQKGEEVQVIAVEPGKNPKNFQTMPGTFQTTTGEIGAIALDFKPGTSGSPIINREGKVVGLYGNGVVTKNGGYVSGIAQTNAEPDGPTPELEEEMFKKRNLTIMDLHPGSGKTRKYLPAIVREAIKRRLRTLILAPTRVVAAEMEEALKGLPIRYQTTATKSEHTGREIVDLMCHATFTMRLLSPVRVPNYNLIIMDEAHFTDPASIAARGYISTRVGMGEAAAIFMTATPPGTADAFPQSNAPIQDEERDIPERSWNSGNEWITDFAGKTVWFVPSIKAGNDIANCLRKNGKKVIQLSRKTFDTEYQKTKLNDWDFVV
>tr|M4KW32|M4KW32_BACIU Choline ABC transporter (ATP-binding protein) OS=Bacillus subtilis XF-1 GN=opuBA PE=4 SV=1 Split=0
MLTLENVSKTYKGGKKAVNNVNLKIAKGEFICFIGPSGCGKTTTMKMINRLIEPSAGKIFIDGENIMDQDPVELRRKIGYVIQQIGLFPHMTIQQNISLVPKLLKWPEQQRKERARELLKLVDMGPEYVDRYPHELSGGQQQRIGVLRALAAEPPLILMDEPFGALDPITRDSLQEEFKKLQKTLHKTIVFVTHDMDEAIKLADRIVILKAGEIVQVGTPDDILRNPADEFVEEFIGKERLIQSSSPDVERVDQIMNTQPVTITADKTLSEAIQLMRQERVDSLLVVDDEHVLQGYVDVEIIDQCRKKANLIGEVLHEDIYTVLGGTLLRDTVRKILKRGVKYVPVVDEDRRLIGIVTRASLVDIVYDSLWGEEKQLAALS
>sp|Q8AWH3|SX17A_XENTR Transcription factor Sox-17-alpha OS=Xenopus tropicalis GN=sox17a PE=2 SV=1 Split=0
MSSPDGGYASDDQNQGKCSVPIMMTGLGQCQWAEPMNSLGEGKLKSDAGSANSRGKAEARIRRPMNAFMVWAKDERKRLAQQNPDLHNAELSKMLGKSWKALTLAEKRPFVEEAERLRVQHMQDHPNYKYRPRRRKQVKRMKRADTGFMHMAEPPESAVLGTDGRMCLESFSLGYHEQTYPHSQLPQGSHYREPQAMAPHYDGYSLPTPESSPLDLAEADPVFFTSPPQDECQMMPYSYNASYTHQQNSGASMLVRQMPQAEQMGQGSPVQGMMGCQSSPQMYYGQMYLPGSARHHQLPQAGQNSPPPEAQQMGRADHIQQVDMLAEVDRTEFEQYLSYVAKSDLGMHYHGQESVVPTADNGPISSVLSDASTAVYYCNYPSA
I got it! :D
OSarr = []
G = 0
for i in range(len(keyLabelrun)):
OSarr.append(keyLabelrun[G])
G += 1
if keyLabelrun[G].count('='):
while keyLabelrun[G].count('OS=') != 1:
G+=1
Maybe next time everyone, thank you!
Due to the syntax, you have to keep track of which part (OS, PE, etc) you're currently parsing. Here's a function to extract the species name from the FASTA header:
def extract_species(description):
species_parts = []
is_os = False
for word in description.split():
if word[:3] == 'OS=':
is_os = True
species_parts.append(word[3:])
elif '=' in word:
is_os = False
elif is_os:
species_parts.append(word)
return ' '.join(species_parts)
You can call it when processing your input file, e.g.:
from Bio import SeqIO
for record in SeqIO.parse('input.fa', 'fasta'):
species = extract_species(record.description)
I am trying to normalize weight units in a string.
Eg:
1.SUCO MARACUJA COM GENGIBRE PCS 300 Millilitre - SUCO MARACUJA COM GENGIBRE PCS 300 ML
2. OVOS CAIPIRAS ANA MARIA BRAGA 10UN - OVOS CAIPIRAS ANA MARIA BRAGA 10U
3. SUCO MARACUJA MAMAO PCS 300 Gram - SUCO MARACUJA MAMAO PCS 300 G
4. SUCO ABACAXI COM MACA PCS 300Milli litre - SUCO ABACAXI COM MACA PCS 300ML
The keyword table is :
unit = ['Kilo','Kilogram','Gram','Milligram','Millilitre','Milli
litre','Dozen','Litre','Un','Und','Unid','Unidad','Unidade','Unidades']
norm_unit = ['KG','KG','G','MG','ML','ML','DZ','L','U','U','U','U','U','U']
I tried to take up these lists as a table but am having difficulty in comparing two dataframes or tables in python.
I tried the below code.
unit = ['Kilo','Kilogram','Gram','Milligram','Millilitre','Milli
litre','Dozen','Litre','Un','Und','Unid','Unidad','Unidade','Unidades']
norm_unit = ['KG','KG','G','MG','ML','ML','DZ','L','U','U','U','U','U','U']
z='SUCO MARACUJA COM GENGIBRE PCS 300 Millilitre'
#for row in mongo_docs:
#z = row['clean_hntproductname']
for x in unit:
for y in norm_unit:
if (re.search(r'\s'+x+r'$',z,re.I)):
# clean_hntproductname = t.lower().replace(x.lower(),y.lower())
# myquery3 = { "_id" : row['_id']}
# newvalues3 = { "$set": {"clean_hntproductname" : 'clean_hntproductname'} }
# ds_hnt_prod_data.update_one(myquery3, newvalues3)
I'm using Python(Jupyter) with MongoDb(Compass). Fetching data from Mongo and writing back to it.
From my understanding you want to:
Update all the rows in a table which contain the words in the unit array, to the ones in norm_unit.
(Disclaimer: I'm not familiar with MongoDB or Python.)
What you want is to create a mapping (using a hash) of the words you want to change.
Here's a trivial solution (i.e. not best solution but would probably point you in the right direction.)
unit_conversions = {
'Kilo': 'KG'
'Kilogram': 'KG',
'Gram': 'G'
}
# pseudo-code
for each row that you want to update
item_description = get the value of the string in the column
for each key in unit_conversion (e.g. 'Kilo')
see if the item_description contains the key
if it does, replace it with unit_convertion[key] (e.g. 'KG')
update the row
Given a dataframe as follows:
firstname lastname email_address \
0 Doug Watson douglas.watson#dignityhealth.org
1 Nick Holekamp nick.holekamp#rankenjordan.org
2 Rob Schreiner rob.schriener#wellstar.org
3 Austin Phillips austin.phillips#precmed.com
4 Elise Geiger egeiger#puracap.com
5 Paul Urick purick#diplomatpharmacy.com
6 Michael Obringer michael.obringer#lashgroup.com
7 Craig Heneghan cheneghan#west-ward.com
8 Kathy Hirst kathleen.hirst#sunovion.com
9 Stefan Bluemmers stefan.bluemmers#grunenthal.com
companyname
0 Dignity Health
1 Ranken Jordan Pediatric Bridge Hospital
2 WellStar Health System
3 Precision Medical Products, Inc.
4 puracap.com
5 Diplomat Specialty Pharmacy
6 Lash Group
7 West-Ward Pharmaceuticals
8 Sunovion Pharmaceuticals
9 GrĂ¼nenthal Group
How could I create possible email addresses using common email patterns as such: firstlast#example.com, first.last#example.com, f.last#example.com, lastF#example.com, first_last#example.com, firstL#example.com, etc.
df['email1'] = df.firstname.str.lower() + '.' + df.lastname.str.lower() + '#' + df.companyname.str.replace('\s+', '').str.lower() + '.com'
print(df['email1'])
Out:
0 doug.watson#dignityhealth.com
1 nick.holekamp#rankenjordanpediatricbridgehospi... --->problematic
2 rob.schreiner#wellstarhealthsystem.com
3 austin.phillips#precisionmedicalproducts,inc..com --->problematic
4 elise.geiger#puracap.com.com --->problematic
...
9995 terry.hanley#kempersportsmanagement.com
9996 christine.marks#geocomp.com
9997 darryl.rickner#doe.com
9998 lalit.sharma#lovelylifestyle.com
9999 parul.dutt#infibeam.com
Some of them seems quite problematic, anyone could help to solve this issue? Thanks a lot.
EDITED:
print(df) after applying #Sajith Herath's solution:
Out:
firstname lastname companyname \
0 Nick Holekamp Ranken ...
email
0 nick. ...
You can use a method to create permutations of username with different separators and define a max length that simplify the domain using company name as follows
import pandas as pd
import random
data = {"firstname":["Nick"],"lastname":["Holekamp"],"companyname":["Ranken \
Jordan Pediatric Bridge Hospital"]}
df = pd.DataFrame(data=data)
max_char = 5
emails = []
def simplify_domain(text):
if len(text)>max_char:
text = ''.join([c for c in text if c.isupper()])
return text.lower()
return text.replace("\s+","").lower()
def username_permutations(first_name,last_name):
# define separators
separators = [".", "_", "-"]
#lower case
combinations = list(map(lambda x:f"{first_name.lower()}{x} \
{last_name.lower()}",separators))
#append a random number to tail
n = random.randint(1, 100)
combinations.extend(list(map(lambda x:f"{x}{n}",combinations)))
return combinations
for index,row in df.iterrows():
usernames = username_permutations(row["firstname"],row["lastname"])
email_permutations = list(map(lambda x: f" \
{x}#{simplify_domain(row['companyname'])}.com",usernames))
emails.append(','.join(email_permutations))
df["email"] = emails
Final result will be nick.holekamp#rjpbh.com,nick_holekamp#rjpbh.com,nick-holekamp#rjpbh.com,nick.holekamp66#rjpbh.com,nick_holekamp66#rjpbh.com,nick-holekamp66#rjpbh.com
you can modify simplify_domain method to validate given string such as removing inc or .com values
I have 2 columns in my data frame - ASIN and keywords . I am trying to groupby ASINs , the groupby is working fine
ASIN keywords
0 B07GFGXMZZ mangalagiri dress materials
1 B07GFGXMZZ pure cotton dress materials for women
2 B07GFGXMZZ suit material party wear for women
3 B076BL4CWB dhakai jamdani
4 B076BL4CWB jamdani
Groupby Code
df.groupby('ASIN').apply(lambda x: x.sum())
Output
but how to add a space for each lambda iteration as of now it is not doing , you can observe the same in output image i have linked
i Tried
df.groupby('ASIN').apply(lambda x:" ".join(x.sum()))
But it didn't work
ASIN
9801321261 98013212619801321261 cane mat with runnercane ...
B008YLNICE B008YLNICEB008YLNICEB008YLNICEB008YLNICEB008YL...
B00P81OJ26 B00P81OJ26B00P81OJ26B00P81OJ26B00P81OJ26B00P81...
B010SZBHEE B010SZBHEEB010SZBHEEB010SZBHEEB010SZBHEEB010SZ...
B01143KAY2 B01143KAY2B01143KAY2B01143KAY2B01143KAY2B01143...
B0157XMD4A B0157XMD4A elephant painted box
B0157XMRJ6 B0157XMRJ6B0157XMRJ6B0157XMRJ6B0157XMRJ6B0157X...