Text Wrangling and Regex

Adapted from Lisa Yan, Will Fithian, Joseph Gonzalez, Deborah Nolan, Sam Lau

Updated by Bella Crouch

Working with text: applying string methods and regular expressions

import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
import seaborn as sns
import zipfile
import pandas as pd

Demo 1: Canonicalizing County Names

states = pd.read_csv("data/county_and_state.csv")
populations = pd.read_csv("data/county_and_population.csv")

# display allows us to view a DataFrame without returning it as an object
display(states)

display(populations)

	County	State
0	De Witt County	IL
1	Lac qui Parle County	MN
2	Lewis and Clark County	MT
3	St John the Baptist Parish	LS

	County	Population
0	DeWitt	16798
1	Lac Qui Parle	8067
2	Lewis & Clark	55716
3	St. John the Baptist	43044

Both of these DataFrames share a "County" column. Unfortunately, formatting differences mean that we can't directly merge the two DataFrames using the "County"s.

states.merge(populations, left_on="County", right_on="County")

	County	State	Population

To address this, we can canonicalize the "County" string data to apply a common formatting.

def canonicalize_county(county_series):
    return (county_series.str.lower()               # lowercase
            .str.replace(' ', '')                   # remove space
            .str.replace('&', 'and')                # replace &
            .str.replace('.', '')                   # remove dot
            .str.replace('county', '')              # remove "county"
            .str.replace('parish', '')              # remove "parish" 
            )

states["County"] = canonicalize_county(states["County"])
populations["County"] = canonicalize_county(populations["County"])

display(states)

display(populations)

	County	State
0	dewitt	IL
1	lacquiparle	MN
2	lewisandclark	MT
3	stjohnthebaptist	LS

	County	Population
0	dewitt	16798
1	lacquiparle	8067
2	lewisandclark	55716
3	stjohnthebaptist	43044

Now, the merge works as expected!

states.merge(populations, left_on="County", right_on="County")

	County	State	Population
0	dewitt	IL	16798
1	lacquiparle	MN	8067
2	lewisandclark	MT	55716
3	stjohnthebaptist	LS	43044

Demo 2: Extracting Log Data

log_fname = 'data/log.txt'
with open(log_fname, 'r') as f:
    log_lines = f.readlines()
log_lines

['169.237.46.168 - - [26/Jan/2014:10:47:58 -0800] "GET /stat141/Winter04/ HTTP/1.1" 200 2585 "http://anson.ucdavis.edu/courses/"\n',
 '193.205.203.3 - - [2/Feb/2005:17:23:6 -0800] "GET /stat141/Notes/dim.html HTTP/1.0" 404 302 "http://eeyore.ucdavis.edu/stat141/Notes/session.html"\n',
 '169.237.46.240 - "" [3/Feb/2006:10:18:37 -0800] "GET /stat141/homework/Solutions/hw1Sol.pdf HTTP/1.1"\n']

Suppose we want to extract the day, month, year, hour, minutes, seconds, and timezone. Looking at the data, we see that these items are not in a fixed position relative to the beginning of the string. That is, slicing by some fixed offset isn't going to work.

log_lines[0][20:31]

'26/Jan/2014'

log_lines[1][20:31]

'/Feb/2005:1'

Instead, we'll need to use some more sophisticated thinking. Let's focus on only the first line of the file.

first = log_lines[0]
first

'169.237.46.168 - - [26/Jan/2014:10:47:58 -0800] "GET /stat141/Winter04/ HTTP/1.1" 200 2585 "http://anson.ucdavis.edu/courses/"\n'

pertinent = first.split("[")[1].split(']')[0] # find the text enclosed in square brackets
day, month, rest = pertinent.split('/')       # split up the date/month/year 
year, hour, minute, rest = rest.split(':')    # split up the hour:minute:second
seconds, time_zone = rest.split(' ')          # split the timezone after the blank space
day, month, year, hour, minute, seconds, time_zone

('26', 'Jan', '2014', '10', '47', '58', '-0800')

This worked, but felt fairly "hacky" – the code above isn't particularly elegant. A much more sophisticated but common approach is to extract the information we need using a regular expression.

Regular Expressions

String Extraction with Regex

Python re.findall returns a list of all extracted matches:

import re

text = "My social security number is 123-45-6789 bro, or actually maybe it’s 321-45-6789.";
pattern = r"[0-9]{3}-[0-9]{2}-[0-9]{4}"
re.findall(pattern, text)

['123-45-6789', '321-45-6789']

Now, let's see vectorized extraction in pandas:

.str.findall returns a Series of lists of all matches in each record.

df_ssn = pd.DataFrame(
    ['987-65-4321',
     'forty',
     '123-45-6789 bro or 321-45-6789',
     '999-99-9999'],
    columns=['SSN'])
df_ssn

	SSN
0	987-65-4321
1	forty
2	123-45-6789 bro or 321-45-6789
3	999-99-9999

# -> Series of lists
pattern = r"[0-9]{3}-[0-9]{2}-[0-9]{4}"
df_ssn['SSN'].str.findall(pattern)

0                 [987-65-4321]
1                            []
2    [123-45-6789, 321-45-6789]
3                 [999-99-9999]
Name: SSN, dtype: object

Extraction Using Regex Capture Groups

The Python function re.findall, in combination with parentheses returns specific substrings (i.e., capture groups) within each matched string, or match.

text = """I will meet you at 08:30:00 pm tomorrow"""       
pattern = ".*(\d\d):(\d\d):(\d\d).*"
matches = re.findall(pattern, text)
matches

[('08', '30', '00')]

# the three capture groups in the first matched string
hour, minute, second = matches[0]

In pandas, we can use .str.extract to extract each capture group of only the first match of each record into separate columns.

# back to SSNs
df_ssn

	SSN
0	987-65-4321
1	forty
2	123-45-6789 bro or 321-45-6789
3	999-99-9999

# Will extract the first match of all groups
pattern_group_mult = r"([0-9]{3})-([0-9]{2})-([0-9]{4})" # 3 groups
df_ssn['SSN'].str.extract(pattern_group_mult)

	0	1	2
0	987	65	4321
1	NaN	NaN	NaN
2	123	45	6789
3	999	99	9999

Alternatively, .str.extractall extracts all matches of each record into separate columns. Rows are then MultiIndexed by original record index and match index.

# -> DataFrame, one row per match
df_ssn['SSN'].str.extractall(pattern_group_mult)

		0	1	2
	match
0	0	987	65	4321
2	0	123	45	6789
	1	321	45	6789
3	0	999	99	9999

Canonicalization with Regex

In regular Python, canonicalize with re.sub (standing for "substitute"):

text = '<div><td valign="top">Moo</td></div>'
pattern = r"<[^>]+>"
re.sub(pattern, '', text)

'Moo'

In pandas, canonicalize with Series.str.replace.

# example dataframe of strings
df_html = pd.DataFrame(['<div><td valign="top">Moo</td></div>',
                   '<a href="http://ds100.org">Link</a>',
                   '<b>Bold text</b>'], columns=['Html'])
df_html

	Html
0	<div><td valign="top">Moo</td></div>
1	<a href="http://ds100.org">Link</a>
2	<b>Bold text</b>

# Series -> Series
df_html["Html"].str.replace(pattern, '', regex=True).to_frame()

	Html
0	Moo
1	Link
2	Bold text

Revisiting Text Log Processing using Regex

Python re version

line = log_lines[0]
display(line)

pattern = r'\[(\d+)\/(\w+)\/(\d+):(\d+):(\d+):(\d+) (.+)\]'
day, month, year, hour, minute, second, time_zone = re.findall(pattern, line)[0] # get first match
day, month, year, hour, minute, second, time_zone

'169.237.46.168 - - [26/Jan/2014:10:47:58 -0800] "GET /stat141/Winter04/ HTTP/1.1" 200 2585 "http://anson.ucdavis.edu/courses/"\n'

('26', 'Jan', '2014', '10', '47', '58', '-0800')

pandas version

df = pd.DataFrame(log_lines, columns=['Log'])
df

	Log
0	169.237.46.168 - - [26/Jan/2014:10:47:58 -0800...
1	193.205.203.3 - - [2/Feb/2005:17:23:6 -0800] "...
2	169.237.46.240 - "" [3/Feb/2006:10:18:37 -0800...

Option 1: Series.str.findall

pattern = r'\[(\d+)\/(\w+)\/(\d+):(\d+):(\d+):(\d+) (.+)\]'
df['Log'].str.findall(pattern)

0    [(26, Jan, 2014, 10, 47, 58, -0800)]
1      [(2, Feb, 2005, 17, 23, 6, -0800)]
2     [(3, Feb, 2006, 10, 18, 37, -0800)]
Name: Log, dtype: object

Option 2: Series.str.extractall

df['Log'].str.extractall(pattern)

		0	1	2	3	4	5	6
	match
0	0	26	Jan	2014	10	47	58	-0800
1	0	2	Feb	2005	17	23	6	-0800
2	0	3	Feb	2006	10	18	37	-0800

Wrangling either of these two DataFrames into a nice format (like below) is left as an exercise for you! You will do a related problem on the homework.

	Day	Month	Year	Hour	Minute	Second	Time Zone
0	26	Jan	2014	10	47	58	-0800
1	2	Feb	2005	17	23	6	-0800
2	3	Feb	2006	10	18	37	-0800

# your code here
...

---

Real World Case Study: Restaurant Data

In this example, we will show how regexes can allow us to track quantitative data across categories defined by the appearance of various text fields.

In this example we'll see how the presence of certain keywords can affect quantitative data:

How do restaurant health scores vary as a function of the number of violations that mention a particular keyword?
(e.g., unclean surfaces, vermin, permits, etc.)

vio = pd.read_csv('data/violations.csv', header=0, names=['bid', 'date', 'desc'])
desc = vio['desc']
vio.head()

	bid	date	desc
0	19	20171211	Inadequate food safety knowledge or lack of ce...
1	19	20171211	Unapproved or unmaintained equipment or utensils
2	19	20160513	Unapproved or unmaintained equipment or utensi...
3	19	20160513	Unclean or degraded floors walls or ceilings ...
4	19	20160513	Food safety certificate or food handler card n...

counts = desc.value_counts()
counts.shape

(14253,)

That's a lot of different descriptions!! Can we canonicalize at all? Let's explore two sets of 10 rows.

counts[:10]

desc
Unclean or degraded floors walls or ceilings                          999
Unapproved or unmaintained equipment or utensils                      659
Inadequately cleaned or sanitized food contact surfaces               493
Improper food storage                                                 476
Inadequate and inaccessible handwashing facilities                    467
Moderate risk food holding temperature                                452
Wiping cloths not clean or properly stored or inadequate sanitizer    418
Moderate risk vermin infestation                                      374
Unclean nonfood contact surfaces                                      369
Food safety certificate or food handler card not available            353
Name: count, dtype: int64

# Hmmm...
counts[50:60]

desc
Unclean or degraded floors walls or ceilings  [ date violation corrected: 11/29/2017 ]              16
Unclean or degraded floors walls or ceilings  [ date violation corrected: 9/19/2017 ]               16
Inadequate HACCP plan record keeping                                                                16
Unclean or degraded floors walls or ceilings  [ date violation corrected: 11/27/2017 ]              15
Unclean or degraded floors walls or ceilings  [ date violation corrected: 12/7/2017 ]               15
Inadequately cleaned or sanitized food contact surfaces  [ date violation corrected: 9/26/2017 ]    14
Unclean or degraded floors walls or ceilings  [ date violation corrected: 11/28/2017 ]              14
Unclean or degraded floors walls or ceilings  [ date violation corrected: 9/6/2017 ]                14
Unapproved or unmaintained equipment or utensils  [ date violation corrected: 9/19/2017 ]           14
Unapproved  living quarters in food facility                                                        13
Name: count, dtype: int64

# Use regular expressions to cut out the extra info in square braces.
vio['clean_desc'] = (vio['desc']
             .str.replace(r'\s*\[.*\]$', '', regex=True)
             .str.strip()       # removes leading/trailing whitespace
             .str.lower())
vio.head()

	bid	date	desc	clean_desc
0	19	20171211	Inadequate food safety knowledge or lack of ce...	inadequate food safety knowledge or lack of ce...
1	19	20171211	Unapproved or unmaintained equipment or utensils	unapproved or unmaintained equipment or utensils
2	19	20160513	Unapproved or unmaintained equipment or utensi...	unapproved or unmaintained equipment or utensils
3	19	20160513	Unclean or degraded floors walls or ceilings ...	unclean or degraded floors walls or ceilings
4	19	20160513	Food safety certificate or food handler card n...	food safety certificate or food handler card n...

# canonicalizing definitely helped
vio['clean_desc'].value_counts().shape

(68,)

vio['clean_desc'].value_counts().head() 

clean_desc
unclean or degraded floors walls or ceilings               3507
moderate risk food holding temperature                     2542
inadequate and inaccessible handwashing facilities         2529
unapproved or unmaintained equipment or utensils           2382
inadequately cleaned or sanitized food contact surfaces    2301
Name: count, dtype: int64

Remember our research question:

How do restaurant health scores vary as a function of the number of violations that mention a particular keyword?
(e.g., unclean surfaces, vermin, permits, etc.)

Below, we use regular expressions and df.assign() (documentation) to method chain our creation of new boolean features, one per keyword.

# use regular expressions to assign new features for the presence of various keywords
# regex metacharacter | 
with_features = (vio
 .assign(is_unclean     = vio['clean_desc'].str.contains('clean|sanit'))
 .assign(is_high_risk = vio['clean_desc'].str.contains('high risk'))
 .assign(is_vermin    = vio['clean_desc'].str.contains('vermin'))
 .assign(is_surface   = vio['clean_desc'].str.contains('wall|ceiling|floor|surface'))
 .assign(is_human     = vio['clean_desc'].str.contains('hand|glove|hair|nail'))
 .assign(is_permit    = vio['clean_desc'].str.contains('permit|certif'))
)
with_features.head()

	bid	date	desc	clean_desc	is_unclean	is_high_risk	is_vermin	is_surface	is_human	is_permit
0	19	20171211	Inadequate food safety knowledge or lack of ce...	inadequate food safety knowledge or lack of ce...	False	False	False	False	False	True
1	19	20171211	Unapproved or unmaintained equipment or utensils	unapproved or unmaintained equipment or utensils	False	False	False	False	False	False
2	19	20160513	Unapproved or unmaintained equipment or utensi...	unapproved or unmaintained equipment or utensils	False	False	False	False	False	False
3	19	20160513	Unclean or degraded floors walls or ceilings ...	unclean or degraded floors walls or ceilings	True	False	False	True	False	False
4	19	20160513	Food safety certificate or food handler card n...	food safety certificate or food handler card n...	False	False	False	False	True	True

EDA

That's the end of our text wrangling. Now let's do some more analysis to analyze restaurant health as a function of the number of violation keywords.

To do so we'll first group so that our granularity is one inspection for a business on particular date. This effectively counts the number of violations by keyword for a given inspection.

count_features = (with_features
 .groupby(['bid', 'date'])
 .sum(numeric_only=True)
 .reset_index()
)
count_features.iloc[255:260, :]

	bid	date	is_unclean	is_high_risk	is_vermin	is_surface	is_human	is_permit
255	489	20150728	5	0	2	3	0	0
256	489	20150807	1	0	0	1	0	0
257	489	20160308	2	2	1	0	1	0
258	489	20160721	2	1	1	1	0	1
259	489	20161220	3	0	1	2	0	0

Check out our new dataframe in action:

count_features[count_features['is_vermin'] > 1].head(5)

	bid	date	is_unclean	is_high_risk	is_vermin	is_surface	is_human	is_permit
255	489	20150728	5	0	2	3	0	0
291	527	20170821	1	1	2	1	1	1
1508	2622	20160526	4	2	2	3	0	0
1573	2721	20150422	2	1	2	1	0	0
1746	2945	20150921	2	1	2	2	2	1

Now we'll reshape this "wide" table into a "tidy" table using a pandas feature called pd.melt (documentation) which we won't describe in any detail, other than that it's effectively the inverse of pd.pivot_table.

Our granularity is now a violation type for a given inspection (for a business on a particular date).

violation_type_df = pd.melt(count_features, id_vars=['bid', 'date'],
            var_name='feature', value_name='num_vios')

# show a particular inspection's results
violation_type_df[(violation_type_df['bid'] == 489) & (violation_type_df['date'] == 20150728)]

	bid	date	feature	num_vios
255	489	20150728	is_unclean	5
12517	489	20150728	is_high_risk	0
24779	489	20150728	is_vermin	2
37041	489	20150728	is_surface	3
49303	489	20150728	is_human	0
61565	489	20150728	is_permit	0

Remember our research question:

How do restaurant health scores vary as a function of the number of violations that mention a particular keyword?
(e.g., unclean surfaces, vermin, permits, etc.)

We have the second half of this question! Now let's join our table with the inspection scores, located in inspections.csv.

# read in the scores
inspection_df = pd.read_csv('data/inspections.csv',
                  header=0,
                  usecols=[0, 1, 2],
                  names=['bid', 'score', 'date'])
inspection_df.head()

	bid	score	date
0	19	94	20160513
1	19	94	20171211
2	24	98	20171101
3	24	98	20161005
4	24	96	20160311

While the inspection scores were stored in a separate file from the violation descriptions, we notice that the primary key in inspections is (bid, date)! So we can reference this key in our join.

# join scores with the table broken down by violation type
violation_type_and_scores = (
    violation_type_df
    .merge(inspection_df, on=['bid', 'date'])
)
violation_type_and_scores.head(12)

	bid	date	feature	num_vios	score
0	19	20160513	is_unclean	1	94
1	19	20160513	is_high_risk	0	94
2	19	20160513	is_vermin	0	94
3	19	20160513	is_surface	1	94
4	19	20160513	is_human	1	94
5	19	20160513	is_permit	1	94
6	19	20171211	is_unclean	0	94
7	19	20171211	is_high_risk	0	94
8	19	20171211	is_vermin	0	94
9	19	20171211	is_surface	0	94
10	19	20171211	is_human	0	94
11	19	20171211	is_permit	1	94

---

Let's plot the distribution of scores, broken down by violation counts, for each inspection feature (is_clean, is_high_risk, is_vermin, is_surface).

# you will learn this syntax next week. Focus on interpreting for now.
sns.catplot(x='num_vios', y='score',
               col='feature', col_wrap=2,
               kind='box',
               data=violation_type_and_scores);

Above we can observe:

• The inspection score generally goes down with increasing numbers of violations, as expected.
• Depending on the violation keyword, inspections scores on average go down at slightly different rates.
• For example, that if a restaurant inspection involved 2 violations with the keyword "vermin", the average score for that inspection would be a little bit below 80.