gistfile1.py

import re

testtext = """1 22 333 444 555 666 777 888 999
999 999 999 testing something else 1bananas"""

# Benford's law applies only to numbers, not to words starting with numbers like '1bananas',
# so let's use a regex to find all the numbers in the text:
numbers = re.findall(r"\b\d+\b", testtext)

# here's how to understand that regular expression:
#
# \b means "word boundary", and matches an empty space at the beginning or end of a word
# \d means "digit", and matches any digit
# + means "one or more"
#
# so we can read \b\d+\b as "a word boundary, followed by one or more integers, followed by
# a word boundary"
#
# numbers is now: ['1', '22', '333', '444', '555', '666', '777', '888', '999', '999', '999', '999']


# now we want to count the number of times each starting digit reoccurs. Let's use the 
# Counter object I introduced in my previous gist:
from collections import Counter
number_counts = Counter()

for n in numbers:
    number_counts[n[0]] += 1


# now that we have our counts, let's calculate the percentage each n represents

# we need to convert this number to a float so that 10/3 == 3.33 instead of 10/3 == 3
total_numbers = float(len(numbers))

# now, the percentage is just (count/total)*100 for each number
number_percentage = [(number_counts[i]/total_numbers) * 100 for i in '123456789']

benfords_law = [30.1, 17.6, 12.5, 9.7, 7.9, 6.7, 5.8, 5.1, 4.6]

difference = [number_percentage[i] - benfords_law[i] for i in range(9)]

for i in range(1, 10):
    print "You had %d %d's (expecting %.2f%%, difference %.2f%%" % (number_counts[str(i)], i, benfords_law[i-1], difference[i-1])