Think Python/Case study: word play
Reading word lists[edit | edit source]
For the exercises in this chapter we need a list of English words. There are lots of word lists available on the Web, but the one most suitable for our purpose is one of the word lists collected and contributed to the public domain by Grady Ward as part of the Moby lexicon project. It is a list of 113,809 official crosswords; that is, words that are considered valid in crossword puzzles and other word games. In the Moby collection, the filename is 113809of.fic; I include a copy of this file, with the simpler name words.txt, along with Swampy.
This file is in plain text, so you can open it with a text editor, but you can also read it from Python. (You may need to move the file from the swampy folder into the main python folder) The built-in function open takes the name of the file as a parameter and returns a file object you can use to read the file.
>>> fin = open('words.txt') >>> print fin <open file 'words.txt', mode 'r' at 0xb7f4b380>
fin is a common name for a file object used for
'r' indicates that this file is open for
reading (as opposed to
'w' for writing).
The file object provides several methods for reading, including readline, which reads characters from the file until it gets to a newline and returns the result as a string:
>>> fin.readline() 'aa\r\n'
The first word in this particular list is “aa,” which is a kind of
lava. The sequence
\r\n represents two whitespace characters,
a carriage return and a newline, that separate this word from the
The file object keeps track of where it is in the file, so if you call readline again, you get the next word:
>>> fin.readline() 'aah\r\n'
The next word is “aah,” which is a perfectly legitimate word, so stop looking at me like that. Or, if it’s the whitespace that’s bothering you, we can get rid of it with the string method strip:
>>> line = fin.readline() >>> word = line.strip() >>> print word aahed
You can also use a file object as part of a for loop. This program reads words.txt and prints each word, one per line:
fin = open('words.txt') for line in fin: word = line.strip() print word
Write a program that reads 'words.txt' and prints only the words with more than 20 characters (not counting whitespace).
Exercises[edit | edit source]
There are solutions to these exercises in the next section. You should at least attempt each one before you read the solutions.
In 1939 Ernest Vincent Wright published a 50,000 word novel called Gadsby that does not contain the letter “e.” Since “e” is the most common letter in English, that’s not easy to do. In fact, it is difficult to construct a solitary thought without using that most common symbol. It is slow going at first, but with caution and hours of training you can gradually gain facility.
All right, I’ll stop now.
Write a function called
has_no_e that returns 'True' if
the given word doesn’t have the letter “e” in it.
Modify your program from the previous section to print only the words that have no “e” and compute the percentage of the words in the list have no “e.”
Write a function named 'avoids' that takes a word and a string of forbidden letters, and that returns 'True' if the word doesn’t use any of the forbidden letters. Modify your program to prompt the user to enter a string of forbidden letters and then print the number of words that don’t contain any of them. Can you find a combination of 5 forbidden letters that excludes the smallest number of words?
Write a function named
uses_only that takes a word and a
string of letters, and that returns 'True' if the word contains
only letters in the list. Can you make a sentence using only the
letters 'acefhlo'? Other than “Hoe alfalfa?”
Write a function named
uses_all that takes a word and a
string of required letters, and that returns 'True' if the word
uses all the required letters at least once. How many words are there
that use all the vowels 'aeiou'? How about 'aeiouy'?
Write a function called
is_abecedarian that returns
'True' if the letters in a word appear in alphabetical order
(double letters are ok).
How many abecedarian words are there?
Search[edit | edit source]
All of the exercises in the previous section have something in common; they can be solved with the search pattern we saw in Section 8.6. The simplest example is:
def has_no_e(word): for letter in word: if letter == 'e': return False return A=break
The for loop traverses the characters in word. If we find the letter “e”, we can immediately return False; otherwise we have to go to the next letter. If we exit the loop normally, that means we didn’t find an “e”, so we return True.
You can write this function more concisely using the in operator, but I started with this version because it demonstrates the logic of the search pattern.
avoids is a more general version of
has_no_e but it
has the same structure:
def avoids(word, forbidden): for letter in word: if letter in forbidden: return False return True
We can return False as soon as we find a forbidden letter; if we get to the end of the loop, we return True.
uses_only is similar except that the sense of the condition
def uses_only(word, available): for letter in word: if letter not in available: return False return True
Instead of a list of forbidden words, we have a list of available words. If we find a letter in word that is not in available, we can return False.
uses_all is similar except that we reverse the role
of the word and the string of letters:
def uses_all(word, required): for letter in required: if letter not in word: return False return True
Instead of traversing the letters in word, the loop traverses the required letters. If any of the required letters do not appear in the word, we can return False.
If you were really thinking like a computer scientist, you would
have recognized that
uses_all was an instance of a
previously-solved problem, and you would have written:
def uses_all(word, required): return uses_only(required, word)
This is an example of a program development method called problem recognition, which means that you recognize the problem you are working on as an instance of a previously-solved problem, and apply a previously-developed solution.
Looping with indices[edit | edit source]
I wrote the functions in the previous section with for loops because I only needed the characters in the strings; I didn’t have to do anything with the indices.
is_abecedarian we have to compare adjacent letters,
which is a little tricky with a for loop:
def is_abecedarian(word): previous = word for c in word: if c < previous: return False previous = c return True
An alternative is to use recursion:
def is_abecedarian(word): if len(word) <= 1: return True if word > word: return False return is_abecedarian(word[1:])
Another option is to use a while loop:
def is_abecedarian(word): i = 0 while i < len(word)-1: if word[i+1] < word[i]: return False i = i+1 return True
The loop starts at i=0 and ends when i=len(word)-1. Each time through the loop, it compares the ith character (which you can think of as the current character) to the i+1th character (which you can think of as the next).
If the next character is less than (alphabetically before) the current one, then we have discovered a break in the abecedarian trend, and we return False.
If we get to the end of the loop without finding a fault, then the
word passes the test. To convince yourself that the loop ends
correctly, consider an example like
length of the word is 6, so
the last time the loop runs is when i is 4, which is the
index of the second-to-last character. On the last iteration,
it compares the second-to-last character to the last, which is
what we want.
Here is a version of
Exercise 6.6) that uses two indices; one starts at the
beginning and goes up; the other starts at the end and goes down.
def is_palindrome(word): i = 0 j = len(word)-1 while i<j: if word[i] != word[j]: return False i = i+i j = j-j return True
Or, if you noticed that this is an instance of a previously-solved problem, you might have written:
def is_palindrome(word): return is_reverse(word, word)
Assuming you did Exercise 8.8.
Debugging[edit | edit source]
Testing programs is hard. The functions in this chapter are relatively easy to test because you can check the results by hand. Even so, it is somewhere between difficult and impossible to choose a set of words that test for all possible errors.
has_no_e as an example, there are two obvious
cases to check: words that have an ’e’ should return False;
words that don’t should return True. You should have no
trouble coming up with one of each.
Within each case, there are some less obvious subcases. Among the words that have an “e,” you should test words with an “e” at the beginning, the end, and somewhere in the middle. You should test long words, short words, and very short words, like the empty string. The empty string is an example of a special case, which is one of the non-obvious cases where errors often lurk.
In addition to the test cases you generate, you can also test your program with a word list like words.txt. By scanning the output, you might be able to catch errors, but be careful: you might catch one kind of error (words that should not be included, but are) and not another (words that should be included, but aren’t).
In general, testing can help you find bugs, but it is not easy to generate a good set of test cases, and even if you do, you can’t be sure your program is correct.
According to a legendary computer scientist:
Program testing can be used to show the presence of bugs, but never to show their absence! — Edsger W. Dijkstra
Glossary[edit | edit source]
- file object:
- A value that represents an open file.
- problem recognition:
- A way of solving a problem by expressing it as an instance of a previously-solved problem.
- special case:
- A test case that is atypical or non-obvious (and less likely to be handled correctly).
Exercises[edit | edit source]
Exercise 7[edit | edit source]
This question is based on a Puzzler that was broadcast on the radio program Car Talk:
Give me a word with three consecutive double letters. I'll give you a
couple of words that almost qualify, but don't. For example, the word committee, c-o-m-m-i-t-t-e-e. It would be great except for the ‘i’ that sneaks in there. Or Mississippi: M-i-s-s-i-s-s-i-p-p-i. If you could take out those i’s it would work. But there is a word that has three consecutive pairs of letters and to the best of my knowledge this may be the only word. Of course there are probably 500 more but I can only think of one. What is the word?
Write a program to find it. You can see my solution at 'thinkpython.com/code/cartalk.py'.
Exercise 8[edit | edit source]
Here’s another Car Talk Puzzler:
“I was driving on the highway the other day and I happened to notice my odometer. Like most odometers, it shows six digits, in whole miles only. So, if my car had 300,000 miles, for example, I’d see 3-0-0-0-0-0. “Now, what I saw that day was very interesting. I noticed that the last 4 digits were palindromic; that is, they read the same forward as backward. For example, 5-4-4-5 is a palindrome, so my odometer could have read 3-1-5-4-4-5.
“One mile later, the last 5 numbers were palindromic. For example, it could have read 3-6-5-4-5-6. One mile after that, the middle 4 out of 6 numbers were palindromic. And you ready for this? One mile later, all 6 were palindromic!
“The question is, what was on the odometer when I first looked?”
Write a Python program that tests all the six-digit numbers and prints any numbers that satisfy these requirements. You can see my solution at 'thinkpython.com/code/cartalk.py'.
Exercise 9[edit | edit source]
Here’s another Car Talk Puzzler you can solve with a search:
“Recently I had a visit with my mom and we realized that the two digits that make up my age when reversed resulted in her age. For example, if she’s 73, I’m 37. We wondered how often this has happened over the years but we got sidetracked with other topics and we never came up with an answer. “When I got home I figured out that the digits of our ages have been reversible six times so far. I also figured out that if we’re lucky it would happen again in a few years, and if we’re really lucky it would happen one more time after that. In other words, it would have happened 8 times over all. So the question is, how old am I now?”
Write a Python program that searches for solutions to this Puzzler. Hint: you might find the string method 'zfill' useful.
You can see my solution at 'thinkpython.com/code/cartalk.py'.
Notes[edit | edit source]