I'm new to Python and Pandas but I try to use Pandas Dataframes to merge two dataframes based on regular expression.
I have one dataframe with some 2 million rows. This table contains data about cars but the model name is often specified in - lets say - a creative way, e.g. 'Audi A100', 'Audi 100', 'Audit 100 Quadro', or just 'A 100'. And the same for other brands. This is stored in a column called "Model". In a second model I have the manufacturer.
Index
Model
Manufacturer
0
A 100
Audi
1
A100 Quadro
Audi
2
Audi A 100
Audi
...
...
...
To clean up the data I created about 1000 regular expressions to search for some key words and stored it in a dataframe called 'regex'. In a second column of this table I save the manufacture. This value is used in a second step to validate the result.
Index
RegEx
Manufacturer
0
.* A100 .*
Audi
1
.* A 100 .*
Audi
2
.* C240 .*
Mercedes
3
.* ID3 .*
Volkswagen
I hope you get the idea.
As far as I understood, the Pandas function "merge()" does not work with regular expressions. Therefore I use a loop to process the list of regular expressions, then use the "match" function to locate matching rows in the car DataFrame and assign the successfully used RegEx and the suggested manufacturer.
I added two additional columns to the cars table 'RegEx' and 'Manufacturer'.
for index, row in regex.iterrows():
cars.loc[cars['Model'].str.match(row['RegEx']),'RegEx'] = row['RegEx']
cars.loc[cars['Model'].str.match(row['RegEx']),'Manufacturer'] = row['Manfacturer']
I learnd 'iterrows' should not be used for performance reasons. It takes 8 minutes to finish the loop, what isn't too bad. However, is there a better way to get it done?
Kind regards
Jiriki
I have no idea if it would be faster (I'll be glad, if you would test it), but it doesn't use iterrows():
regex.groupby(["RegEx", "Manufacturer"])["RegEx"]\
.apply(lambda x: cars.loc[cars['Model'].str.match(x.iloc[0])])
EDIT: Code for reproduction:
cars = pd.DataFrame({"Model": ["A 100", "A100 Quatro", "Audi A 100", "Passat V", "Passat Gruz"],
"Manufacturer": ["Audi", "Audi", "Audi", "VW", "VW"]})
regex = pd.DataFrame({"RegEx": [".*A100.*", ".*A 100.*", ".*Passat.*"],
"Manufacturer": ["Audi", "Audi", "VW"]})
#Output:
# Model Manufacturer
#RegEx Manufacturer
#.*A 100.* Audi 0 A 100 Audi
# 2 Audi A 100 Audi
#.*A100.* Audi 1 A100 Quatro Audi
#.*Passat.* VW 3 Passat V VW
# 4 Passat Gruz VW
Related
This question already has answers here:
Filter pandas DataFrame by substring criteria
(17 answers)
Closed 2 years ago.
I have the following dataframe:
pd.DataFrame({
'Code': ['XAW', 'PAK', 'I', 'QP', 'TOPZ', 'XAW', 'APOL'],
'Name': ['George Truck', 'Fred Williams', 'Jessica Weir', 'Tony P.', 'John Truck', 'Liz Moama', 'Emily Truck'],
'Color': ['Blue', 'Green', 'Green', 'Red', 'Pink', 'Blue', 'Pink']
})
Code Name Color
0 XAW George Truck Blue
1 PAK Fred Williams Green
2 I Jessica Weir Green
3 QP Tony P. Red
4 TOPZ John Truck Pink
5 XAW Liz Moama Blue
6 APOL Emily Truck Pink
Given a keyword, such as 'blue', I would like to retrieve the following rows:
0 XAW George Paul Blue
5 XAW Liz Moama Blue
The search can contain multiple keywords, for example, 'truck pink' would return:
4 TOPZ John Truck Pink
6 APOL Emily Truck Pink
Imagine that this dataframe has half a million rows and a few extra columns. Is there a fast way I can query the entire dataframe for specific keywords?
With search string s = 'truck pink', set up a search column:
t = (df['Name'] + ' ' + df['Color']).str.lower()
I force everything to lower case, because your search example doesn't seem to be case-sensitive. If you have dynamic search inputs, also force the search field to lower case. Then do searches for contains like so:
d = {}
for i in s.split(' '):
d[i] = t.str.contains(i, na=False)
I pass na=False because otherwise, Pandas will fill NA in cases where the string column is itself NA. We don't want that behaviour. The complexity of the operation increases rapidly with the number of search terms. Also consider changing this function if you want to match whole words, because contains matches sub-strings.
Regardless, take results and reduce them with bit-wise 'and'. You need two imports:
from functools import reduce
from operator import and_
df[reduce(and_, d.values())]
And thus:
Code Name Color
4 TOPZ John Truck Pink
6 APOL Emily Truck Pink
I would like to create an aspect analysis from user reviews. The reviews contain various aspects and therefore the reviews need to be separated into sentences. I save the data in a pandas dataframe and separate the sentences with the nltk library.
I put the separate sentences in a list that I want to format into a dataframe and connect to the original dataframe. However, I get an error. Instead of an extra column, I get 19 new columns. (the individual sentences are not stored in a cell, I think every single sentence gets their own column) I also tested itertools but I also get a wrong record.
Can someone help me to get the right format?
I would like to have a new dataframe which looks like that:
U_REVIEW | SENTENCES
--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Im a Sentence. Iam another Sentence in a Row. |[u'Im a Sentence', u'Iam another Sentence in a Row.']
Here we go, next Sentence. Blub, more blubs. |[u"Here weg o, next Sentence.", u'Blub, more blubs.']
Once again, more Sentence. And some other information. The Restaurant was ok, but not awesome.|[u"Once again, more Sentence.", u'And some other information.',u’The Restaurant was ok, but not awesome.’]
That’s how my code looks like:
ta = ta[['U_REVIEW']]
Output:
U_REVIEW
Im a Sentence. Iam another Sentence in a Row.
Here we go, next Sentence. Blub, more blubs.
Once again, more Sentence. And some other information. The Restaurant was ok, but not awesome.
# the empty lists
sentences = []
ss = []
for sentence in ta['U_REVIEW']:
# seperates the review into sentence
sentence = sent_tokenize(sentence)
sentences.append(sentence)
test = itertools.chain(sentences)
#new dataframe to add the Sentences
df2 = pd.DataFrame(sentences)
#create Column
cols2 = ['REVIEW_SENTENCES']
# bring the two dataframes together
df2 = pd.DataFrame(sentences, columns=cols2)
Output of senteces:
[[u'Im a Sentence', u'Iam another Sentence in a Row.'],[u"Here weg o, next Sentence.", u'Blub, more blubs.'],[u"Once again, more Sentence.", u'And some other information.',u’The Restaurant was ok, but not awesome.’]]
Output of test:
<itertools.chain object at 0x000000001316DC18>
Output and Information of the new Dataframe df2:
AssertionError: 1 columns passed, passed data had 19 columns
U_REVIEW | 0 | 1 | 2 ...
-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Im a Sentence. Iam another Sentence in a Row. |Im a Sentence |Iam another Sentence in a Row. |
Here we go, next Sentence. Blub, more blubs. |Here we go, next Sentence.|Blub, more blubs. |
Once again, more Sentence. And some other information. The Restaurant was ok, but not awesome.|Once again, more Sentence.|And some other information. |The Restaurant was ok, but not awesome.
Here is a Testset of a Dataframe:
import pandas as pd
ta = pd.DataFrame( ['Im a Sentence. Iam another Sentence in a Row','Here we go, next Sentence. Blub, more blubs.','Once again, more Sentence. And some other information. The Restaurant was ok, but not awsome.'])
ta.columns =['U_REVIEW']
try this I have done it in python 3.5 I think it should work for 2.5 also:
In [45]: df = pd.DataFrame(ta.U_REVIEW.str.split('.',expand=True).replace('',np.nan).fillna(np.nan).values.flatten()).dropna()
In [46]: df
Out[46]:
0
0 Im a Sentence
1 Iam another Sentence in a Row
4 Here we go, next Sentence
5 Blub, more blubs
8 Once again, more Sentence
9 And some other information
10 The Restaurant was ok, but not awsome
is this what you want:
ta.U_REVIEW.str.split('.',expand=True)
Out[50]:
0 1 \
0 Im a Sentence Iam another Sentence in a Row
1 Here we go, next Sentence Blub, more blubs
2 Once again, more Sentence And some other information
2 3
0 None None
1 None
2 The Restaurant was ok, but not awsome
or
In [52]: ta.U_REVIEW.str.split('.').apply(list)
Out[52]:
0 [Im a Sentence, Iam another Sentence in a Row]
1 [Here we go, next Sentence, Blub, more blubs, ]
2 [Once again, more Sentence, And some other in...
Name: U_REVIEW, dtype: object
I currently searching for a method in R which let's me match/merge two data frames. Helas both of these data frames contain non optimal data. They can have certain abbreviations of even typo's in them. Therefore I would like to define a list for each abbreviation and if a string contains one of those elements. If the original entries don't match, R should check if any of the other options of the abbreviation has a match. To illustrate: the name of a company could end with "Limited" but also with "Ltd." of "Ltd" etc.
EXAMPLE
Data
The Original "Address" file contains:
Company name Address
Deloitte Ltd. New York
Coca-Cola New York
Tesla ltd California
Microsoft Limited Washington
Would have to be merged with "EnterpriseNrList"
Company name EnterpriseNumber
Deloitte Ltd. 221
Coca-Cola 334
Tesla ltd 725
Microsoft Limited 127
So the abbreviations should work in "both directions". That's why I said, if R recognises any of the abbreviations, R should try to match all of them.
All of the matches should be reported as the return.
Therefore I would make up a list "Abbreviations" for each possible abbreviation
Limited.
limited
Ltd.
ltd.
Ltd
ltd
Questions
1) Would this be a good method, or would there be a more efficient way?
2) How can I check a list against a list of possible abbreviations (step 1, see below), sort of a containsx from excel?
3) How could I make up a list that replaces for the entries that do not match the abbreviation with all other abbreviatinos (step 2, see below)?
Thoughts for solution
Step 1
As I am still very new to this kind of work, I was thinking the following: use a regex expression to filter out wether a string contains any of the abbreviation options and create a list which will then contain either -1 if no match could be found and >0 if match is found. The no pattern matching can already be matched against the "Address" list. With the other entries I continue to step 2.
In this step I don't really know how to check against a list of options ("Abbreviations" list).
Step 2
Next I would create a list with the matches from step 1 and rbind together all options. In this step I don't really know to I could create a list that combines f.e. Coca-Cola with all it's possible abbreviations.
Coca-Cola Limited
Coca-Cola Ltd.
Coca-Cola Ltd
etc.
Step 3
Lastly I would match/merge this more complete list of companies again with the original "Data" list. With the introduction of step 2 I thought It might be a bit easier on the required computing power, as the original list is about 8000 rows.
I would go in a different approach, fixing the tables first before the merge.
To fix with abreviations, I would use a regex, case insensitive, the final dot being optionnal, I start with a list of 'Normal word' = vector of abbreviations.
abbrevs <- list('Limited'=c('Limited','Ltd'),'Incorporated'=c('Incorporated','Inc'))
The I build the corresponding regex (alternations with an optional dot at end, the case will be ignored by parameter in gsub and agrep later):
regexes <- lapply(abbrevs,function(x) { paste0("(",paste0(x,collapse='|'),")[.]?") })
Which gives:
$Limited
[1] "(Limited|Ltd)[.]?"
$Incorporated
[1] "(Incorporated|Inc)[.]?"
Now we have to apply each regex to the company.name column of each df:
for (i in seq_along(regexes)) {
Address$Company.name <- gsub(regexes[[i]], names(regexes[i]), Address$Company.name, ignore.case=TRUE)
Enterprise$Company.name <- gsub(regexes[[i]], names(regexes[i]), Enterprise$Company.name, ignore.case=TRUE)
}
This does not take into account typos. Here you'll need to work on with agrepor adist to manage it.
Result for Address example data set:
> Address
Company.name Address
1 Deloitte Limited New York
2 Coca-Cola New York
3 Tesla Limited California
4 Microsoft Limited Washington
Input data used:
Address <- structure(list(Company.name = c("Deloitte Ltd.", "Coca-Cola",
"Tesla ltd", "Microsoft Limited"), Address = c("New York", "New York",
"California", "Washington")), .Names = c("Company.name", "Address"
), class = "data.frame", row.names = c(NA, -4L))
Enterprise <- structure(list(Company.name = c("Deloitte Ltd.", "Coca-Cola",
"Tesla ltd", "Microsoft Limited"), EnterpriseNumber = c(221L,
334L, 725L, 127L)), .Names = c("Company.name", "EnterpriseNumber"
), class = "data.frame", row.names = c(NA, -4L))
I would say that the answer depends on whether you have a list of abbreviations or not.
If you have one, you could just look which element of your list contains an abbreviation with grep or greplfunctions. (grep return all indexes that have a matching pattern whereas grepl returns a logical vector).
Also, use the ignore.case= TRUE parameter of these function, so you don't have to try all capitalized/lowercase possibilities.
If you don't have such a list, my first guest would be to extract the first "word" of each company (I would guess that there is a single "Deloitte" company, and that it is "Deloitte Ltd"). You can do so with:
unlist(strsplit(CompanyNames,split = " "))
If you wanted to also correct for typos, this is more a question of string distance.
Hope that it helped!
I have a frequency table of words which looks like below
> head(freqWords)
employees work bose people company
1879 1804 1405 971 959
employee
100
> tail(freqWords)
youll younggood yoyo ytd yuorself zeal
1 1 1 1 1 1
I want to create another frequency table which will combine similar words and add their frequencies
In above example, my new table should contain both employee and employees as one element with a frequency of 1979. For example
> head(newTable)
employee,employees work bose people
1979 1804 1405 971
company
959
I know how to find out similar words (using adist, stringdist) but I am unable to create the frequency table. For instance I can use following to get a list of similar words
words <- names(freqWords)
lapply(words, function(x) words[stringdist(x, words) < 3])
and following to get a list of similar phrases of two words
lapply(words, function(x) words[stringdist2(x, words) < 3])
where stringdist2 is follwoing
stringdist2 <- function(word1, word2){
min(stringdist(word1, word2),
stringdist(word1, gsub(word2,
pattern = "(.*) (.*)",
repl="\\2,\\1")))
}
I do not have any punctuation/special symbols in my words/phrases. (I do not know a lot of R; I created stringdist2 by tweaking an implementation of adist2 I found here but I do not understand everything about how pattern and repl works)
So I need help to create new frequency table.
I've found numerous examples of how to match and update an entire list with one pattern and one replacement, but what I am looking for now is a way to do this for multiple patterns and multiple replacements in a single statement or loop.
Example:
> print(recs)
phonenumber amount
1 5345091 200
2 5386052 200
3 5413949 600
4 7420155 700
5 7992284 600
I would like to insert a new column called 'service_provider' with /^5/ as Company1 and /^7/ as Company2.
I can do this with the following two lines of R:
recs$service_provider[grepl("^5", recs$phonenumber)]<-"Company1"
recs$service_provider[grepl("^7", recs$phonenumber)]<-"Company2"
Then I get:
phonenumber amount service_provider
1 5345091 200 Company1
2 5386052 200 Company1
3 5413949 600 Company1
4 7420155 700 Company2
5 7992284 600 Company2
I'd like to provide a list, rather than discrete set of grepl's so it is easier to keep country specific information in one place, and all the programming logic in another.
thisPhoneCompanies<-list(c('^5','Company1'),c('^7','Company2'))
In other languages I would use a for loop on on the Phone Company list
For every row in thisPhoneCompanies
Add service provider to matched entries in recs (such as the grepl statement)
end loop
But I understand that isn't the way to do it in R.
Using stringi :
library(stringi)
recs$service_provider <- stri_replace_all_regex(str = recs$phonenumber,
pattern = c('^5.*','^7.*'),
replacement = c('Company1', 'Company2'),
vectorize_all = FALSE)
recs
# phonenumber amount service_provider
# 1 5345091 200 Company1
# 2 5386052 200 Company1
# 3 5413949 600 Company1
# 4 7420155 700 Company2
# 5 7992284 600 Company2
Thanks to #thelatemail
Looks like if I use a dataframe instead of a list for the phone companies:
phcomp <- data.frame(ph=c(5,7),comp=c("Company1","Company2"))
I can match and add a new column to my list of phone numbers in a single command (using the match function).
recs$service_provider <- phcomp$comp[match(substr(recs$phonenumber,1,1), phcomp$ph)]
Looks like I lose the ability to use regular expressions, but the matching here is very simple, just the first digit of the phone number.