Section 7.3 Dictionaries and Files

Data: romeo.txt

But soft what light through yonder window breaks
It is the east and Juliet is the sun
Arise fair sun and kill the envious moon
Who is already sick and pale with grief

One of the common uses of a dictionary is to count the occurrence of words in a file with some written text. Let’s start with a very simple file of words taken from the text of Romeo and Juliet.

For the first set of examples, we will use a shortened and simplified version of the text with no punctuation. Later we will work with the text of the scene with punctuation included.

But soft what light through yonder window breaks
It is the east and Juliet is the sun
Arise fair sun and kill the envious moon
Who is already sick and pale with grief

We will write a Python program to read through the lines of the file, break each line into a list of words, and then loop through each of the words in the line and count each word using a dictionary.

You will see that we have two for loops. The outer loop is reading the lines of the file and the inner loop is iterating through each of the words on that particular line. This is an example of a pattern called nested loops because one of the loops is the outer loop and the other loop is the inner loop.

Because the inner loop executes all of its iterations each time the outer loop makes a single iteration, we think of the inner loop as iterating “more quickly” and the outer loop as iterating more slowly.

The combination of the two nested loops ensures that we will count every word on every line of the input file.

Checkpoint 7.3.1.

Q-2: A(n) _________ is a pattern consisting of two for loops, where one of the loops is the outer loop and the other is the inner loop.

In our else statement, we use the more compact alternative for incrementing a variable. counts[word] += 1 is equivalent to counts[word] = counts[word] + 1. Either method can be used to change the value of a variable by any desired amount. Similar alternatives exist for -=, *=, and /=.

Checkpoint 7.3.2.

Q-3: Which of these answers (if any) will increment a key’s value in the dictionary counts by 1?

counts[word] = counts[word] + 1
Almost...try again!
counts[word] += 1
Almost...try again!
None of these options increments the key’s value by 1
Try again! Remember you can increment by either adding 1 or doing += 1.
Both a and b increment the key’s value by 1
Correct! Both of these methods are correct ways to increment a key’s value by 1.

When we run the program, we see a raw dump of all of the counts in unsorted hash order. (the romeo.txt file is available at www.py4e.com/code3/romeo.txt)

python count1.py
Enter the file name: romeo.txt
{'and': 3, 'envious': 1, 'already': 1, 'fair': 1,
'is': 3, 'through': 1, 'pale': 1, 'yonder': 1,
'what': 1, 'sun': 2, 'Who': 1, 'But': 1, 'moon': 1,
'window': 1, 'sick': 1, 'east': 1, 'breaks': 1,
'grief': 1, 'with': 1, 'light': 1, 'It': 1, 'Arise': 1,
'kill': 1, 'the': 3, 'soft': 1, 'Juliet': 1}

It is a bit inconvenient to look through the dictionary to find the most common words and their counts, so we need to add some more Python code to get us the output that will be more helpful.

Checkpoint 7.3.3.

Construct a block of code to read through the lines of a file, break each line into a list of words, and then loop through each of the words in the line and count each word using a dictionary.

file_to_read = "textfile.txt"
---
with open(file_to_read, "r") as filename:
---
with open(file_to_read, "w") as filename:  #distractor
---
 lines = filename.readlines()
---
 word_count = {}
---
 word_count = dictionary()  #distractor
---
 for line in lines:
---
 for word in lines.split()  #distractor
---
  for word in lines.split():
---
   if word not in word_count.keys():
---
    word_count[word] = 0
---
   counts[word] += 1
---
print(word_count)

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