CS 150 - Lab 10

The purpose of this lab is to:

Implement a sorting algorithm.
Learn about Sets and Dictionaries.
Practice reading from files and string manipulation.
Do fun stuff with words.

Getting Started

Dictionaries

As we've seen in class, a dictionary, like a list, is one of the built-in data structures supported by Python for representing collections of data. The entries in a dictionary are key-value pairs. A dictionary key can be any immutable type (typically a number or a string), while a dictionary value can be anything at all. We think of a dictionary as something that maps keys to values. In a traditional dictionary, the keys are words, and the values are their definitions. In a Python dictionary, your key-value pairs could be words and definitions, or usernames and passwords, or phone numbers and names.

Unlike a list, dictionaries are not ordered -- they are just collections of key-value pairs. So there is no element 0, element 1, etc. Instead, values in a dictionary are indexed by their keys. If you print a dictionary, the elements will be listed in a seemingly random order. One advantage of using a dictionary is that testing membership or looking up a value in a dictionary is very fast. To evaluate the boolean expression a in myList, Python needs to look through the whole list, so this gets slower and slower as the list grows. Evaluating a in myDictionary, however, is very fast, even for very large dictionaries. Plus, sometimes it's just easier to have values indexed by arbitrary keys.

Here are some examples of syntax involving dictionaries.

  score = {}                        # set score to be an empty dictionary
  score = {"kirk": 17, "spock": 4}  # set score to be a dictionary with two key-value pairs
  score["spock"]                    # 4
  score["bones"]                    # error (key not found)
  "bones" in score                  # False
  score["bones"] =  42              # adds key "bones" with value 42
  score["spock"] =  18              # updates value for key "spock" to 18
  "bones" in score                  # True
  del score["kirk"]                 # remove key "kirk" and its value
  list(score.keys())                # returns a list of the keys
  len(score)                        # 2 (number of keys in score)
  for k in score :                  # iterates over the keys in score

Distilling Text

20 points, individual

Consider the following text:

   Question:
   Whether nobler mind suffer
   Slings arrows outrageous fortune
   Take arms against sea troubles
   By opposing them.

I suspect many of you recognize this as a condensed version of the first few lines of Hamlet's famous "To be, or not to be" soliloquy. The original unedited text is:

   To be, or not to be -- that is the question:
   Whether 'tis nobler in the mind to suffer
   The slings and arrows of outrageous fortune
   Or to take arms against a sea of troubles
   And by opposing end them.

The former was produced by finding the 30 most commonly used words in the speech (only some of which is shown above) and removing them. Your first challenge is to write a program called distill.py that prompts the user for the name of a text file and a number n, and prints the contents of that text file with the n most common words removed.

Program Outline

The details of implementing a solution are up to you, but here is a suggested outline of how to approach the problem. As usual, think about the 6 steps of program development, and test each piece as you go.

Read in a text document. Here are two for you to try: hamlet.txt, lincoln.txt.
Create a dictionary wordcount to keep track of the word counts for the given text file. That is, the keys in your dictionary should be strings (the words) and the value for a given key should indicate how often that word appeared.
Once you've built the dictionary, implement your own version of insertion sort to build a list called sortcount of word-count pairs, sorted by word frequency, with the most common words first. For example, the first portion of the sortcount for hamlet.txt looks like [('the', 20), ('to', 15), ('of', 15), ('and', 12), ('that', 7),....
Create a list commonwords by dropping all but the first n elements of sortcount and only copying the word part of each pair (we don't need the counts any more). Continuing with the previous example, commonwords would become ['the', 'to', 'of', 'and', 'that',....
Reopen the original text document, and print each word so long as it doesn't appear in commonwords.

Suggestions and Tips

Recall that to read in a text file as a single string, you can use


   textfile = open('hamlet.txt','r')
   textstring = textfile.read()

The string function split() returns a list of all the "words" in a string, where "words" are substrings that don't contain whitespace.
Since you want to identify all the strings that look like "for", "For", "for,", etc. as the same string, it may be helpful to write a function called cleanstring that takes in a string, and returns a version of that string with lowercase letters and all punctuation removed. Testing if a character is punctuation isn't too hard: you can write if ch in ".,:;'\"\n", for example, to test whether ch is a comma, period, colon, semi-colon, single quote, double quote, or newline. Of course, there may be other punctuation you want to check for as well. To make all alphabetic characters lowercase, use the string function lower().
When you're deciding whether to print a word or not, you'll want to look at the "cleaned" version, but you may want to print the original version.
Try to preserve newlines, punctuation and capitalization in your output. The exact details of how to handle these cases is left up to you (for example, you might want to capitalize the first character of each new line, or drop words that don't contain any letters, such as "--"). In order to preserve newlines, it may be helpful to clean and print a line at a time. Remember you can use the for l in textfile syntax to get each line of a file, one at a time.
Keep in mind that your output may differ slightly due to ties in word counts.

Sets

A set is another built-in data structures supported by Python for the mathematical notion of a set, i.e. a collection of elements. Unlike a dictionary, the elements in a set don't have values associated with them. You could simulate a set using a dictionary, but adding a key for each element, and setting that key's value to something arbitrary, like 0, or an empty string, or none. That said, if you don't have data associated with each element, and simply whant to keep track of a set of items, using a set is the way to go.

Like dictionarys (and unlike lists), sets are not ordered, but testing membership and addinging or removing elements is very fast. Sets do not store duplicate elements: adding an element to a set that already contains that element has no effect.

Here are some examples of syntax involving sets.

  team = set()                       # makes a set with 0 elements
  team = {"kirk", "spock"}           # makes a set with 2 elements
  len(team)                          # 2 
  team.add("bones")                  # adds "bones" to team
  team.remove("kirk")                # removes "kirk" from team
  for p in team :                    # iterates through elements of team
  "bones" in team                    # True
  "malcolm" in team                  # False
  "river" not in team                # True

Anagrams

20 points, partner

An anagram is just a rearrangement of the letters in the word to form another word or words. For example, if you rearrange the letters in

   oberlin student

you can get

   let none disturb

   intends trouble

and many many more. But I'm fairly confident that those are probably the best ones.

For this part of the assignment, you'll be writing a program called anagrams.py to generate your own anagrams. To decide which anagrams are at least plausibly interesting, your program will have to decide which strings are legitimate words. Your program should prompt the user for a file containing a word list, and a word for anagramming.

Once we get the basic functionality down, we'll add a few other features. For example, we'd like to allow the user to specify that they are only interested in using words of length 4 or greater, or anagrams containing at most 3 words.

Program Outline

Broadly speaking, the steps you'll follow will include the following.

Read in a text document containing a word list. Here are three: words1.txt, words2.txt. The first is very small, just for testing purposes. The second contains about 4000 common words. Even for relatively short strings, we'll need to use some optimizations if we want to generate anagrams using that word list.
Build a set words containing each word from the text file. Since we have a lot of words, using a set rather than a list will save us a lot of time when testing membership (which is basically all we'll be using it for).
Create a function called contains(s, word) which returns two values. The first value should be a boolean indicating whether the string s contains the letters necessary to spell word. If the answer is True, the second value should be what remains of s after the letters in word have been removed. If the answer is False, the second value returned should just be an empty string. For example,
```
	 contains("zombiepig", "bozo")       # returns False, ""
	 contains("zombiepig", "biz")        # returns True, "omepig"
	
```
Create a recursive function called grams(s, words, sofar) that takes in a string s, a set of words words, and a list of words sofar. This function finds all anagrams of s using elements found in words. Each of these anagrams is printed, along with the words in sofar, on its own line.

You might be wondering why we're passing around the variable sofar. Indeed, when we want to find the anagrams of a string given by the user, we'll pass in an empty list. However, that list will be critical for making use of recursion. Let's look at an example to see why. Suppose we want to find anagrams of
```
   robopirate 
```
We'll look through our wordlist for words that are contained in this string. The string "cat" doesn't appear in "robopirate", but "air" does. So one thing our function call will do is begin looking through the remainder of "robopirate" with "air" removed, looking for further anagrams. That is, it'll continue to look for strings contained in
```
   robopte 
```
Our list includes "bro", which is contained in "robopte", so another recursive call will be made on the remains, namely "opte". Our wordlist contains "poet", leaving us with an empty string. At this point we've used up all the letters in the string, so we have an anagram, namely
```
 air bro poet 
```
Unfortunately, if we want to print our anagram, we're in trouble, since we haven't kept a record of the previous words we found. (Could we have just printed words as we found them?) That's where sofar comes in. This list will track the words we've found so far in this particular branch of the recursion. That is,
```
 grams("robopirate", words, []) 
```
will call (among other things)
```
 grams("robopte", words, ["air"]) 
```
which in turn calls
```
 grams("opte", words, ["air", "bro"]) 
```
which in turn calls
```
 grams("", words, ["air", "bro", "poet"]) 
```
which can now print the complete anagram.

With this in mind, we're ready to describe the overall structure of this function. We loop over every word w in our wordlist. For each word w that's found in our string s, we make a recursive call on the remainder of s, and with a new list, equal to the current list with w added on. Make sure you are not changing sofar. (Why does it need to be a new list? I.e., why would sofar.append(w) be the wrong thing to do?) If we make a call where s is the empty string, we can just print the contents of sofar.

Suggestions and Tips

When trying to determine whether one string s contains a word w, the string replace function is very useful. In particular,
```
s.replace(ch,'',1) 
```
creates a new string that is identical to s except the first occurence (parameter 3) of ch (parameter 1) is replaced by the empty string (parameter 2).
When printing a list of strings, the string join function is pretty handy:
```
" ".join(L)
```
returns a new string containing every element of the list L, glued together with a space. Of course, you could join the elements with any string, but a space makes the most sense here.

Test output

If you print all strings from words1.txt that are contained in "robopirate", you should get the following:

or bro rat bat air ape poet poor ripe taboo orbit

Note that this doesn't contain "rabbit" ("robotpirate" only has one "b"). If you run your program using words1.txt for your word list on the string "robopirate", you should get

   or ape orbit
   or orbit ape
   bro air poet
   bro poet air
   air bro poet
   air poet bro
   ape or orbit
   ape orbit or
   poet bro air
   poet air bro
   orbit or ape
   orbit ape or

It doesn't matter if your output is in another order.

Improvements

Once that's working, add in the following optimization and improvements:

Things slow down a lot when we have longer word lists. One nice optimization makes use of the following observation: if the user's string (s) doesn't contain a particular word (w), then no remainder of s will contain w either. So instead of iterating through the set of all words at each step of the recursion, we need only iterate through the "plausible" words.
Add a "preprocessing" step: Instead of adding every string found in the word file to the set words, only add those that are contained in the user's input string. Fortunately you already have created a function that can help you out here.
Let the user specify a minimum length of words allowable in their anagrams.
Let the user specify a maximum number of words allowable in their anagrams.
Have all parameters (the string, the word file, min length, and max words) input via the command line. Look back at lab 08 if you need a reminder on how to do this. Use exceptions to handle invalid user input.

Handin

If you followed the Honor Code in this assignment, insert a paragraph attesting to the fact within one of your .py files.

I affirm that I have adhered to the Honor Code in this assignment.

You now just need to electronically handin all your files. As a reminder

 
     % cd             # changes to your home directory
     % cd cs150       # goes to your cs150 folder
     % handin         # starts the handin program
                      # class is 150
                      # assignment is 10
                      # file/directory is lab10
     % lshand         # should show that you've handed in something

You can also specify the options to handin from the command line

 
     % cd ~/cs150     # goes to your cs150 folder
     % handin -c 150 -a 10 lab10

File Checklist

You should have submitted the following files:

   distill.py
   anagrams.py

If your partner is submitting, they will submit anagrams.py with both of your names, and you will submit a README with their name.