# Simple Censor Program

I made a function to censor a certain word from a piece of text. Is there anything I should fix or do more efficiently? What are some other approaches I could have taken to detecting words and censoring them?(This took me a lot of time to brainstorm)

Also, I would appreciate it if you told me how you brainstorm how to write a program.

def detect_word_start(text,word):
count = 0
text_count = 0
for x in text:
text_count+=1
if x == word[count]:
count+=1
if count == len(word):
return text_count - len(word) + 1
else:
count = 0

def censor(text,word):
word_start = detect_word_start(text,word)
censored_text = ''
count = 0
for x in text:
count += 1
if count >= word_start and count < word_start+len(word):
censored_text += '*'
else:
censored_text += x
print(censored_text)

• You could have a look here if you're searching for ideas :) – Grajdeanu Alex Aug 3 '18 at 10:38
• oh damn, I just realized I could make this program way simpler with .split() – zoklev Aug 3 '18 at 12:08
• That's cl***ic. – Joshua Aug 3 '18 at 18:20

## Bugs

The censor() function prints its progress as a triangle of text, as censored_text lengthens one character at a time. Did you accidentally indent the print() statement one level too far? In any case, it would be better practice to return its output, so that the caller can choose to print() it instead (or do something else with it).

"Censoring" implies that all appearances of the word should be replaced. However, your censor() function only replaces the first occurrence, since count is never reset to 0.

Your detect_word_start(text, word) function is basically text.index(word), with some differences:

• If the word is not found, it returns None, whereas text.index(word) would raise a ValueError exception. Since the censor() function doesn't handle the possibility that the word never appears in the text, your code could crash. (I would expect the censor() function to return the text unmodified, if there is nothing to redact.)

• It returns a one-based index, which is very unconventional in Python, where indexes start from 0. (Your censor() function also counts characters starting from 1, so at least you are consistent.)

• It misses some possible results because it makes one pass through the text, and will not backtrack. For example, detect_word_start('coconut', 'con') and detect_word_start('nonoperational', 'nope') both return None.

That might be acceptable behaviour, if you are only searching for complete space-delimited words. However, from the way the code was written, it appears that this behaviour was unintended.

## More expressive and efficient code

Counting loops are usually better written using enumerate().

Python supports chained comparisons:

Comparisons can be chained arbitrarily, e.g., x < y <= z is equivalent to x < y and y <= z

Here is the same code, preserving your 1-based indexing convention and other censorship bugs:

def censor(text, word):
word_start = detect_word_start(text, word)
censored_text = ''
for i, char in enumerate(text, 1):
if word_start <= i < word_start + len(word):
censored_text += '*'
else:
censored_text += x
return censored_text


We can save a few lines using a conditional expression. (I know, it looks worse, but there's a reason that you'll see…)

def censor(text, word):
word_start = detect_word_start(text, word)
censored_text = ''
for i, char in enumerate(text):
censored_text += '*' if (word_start <= i < word_start + len(word)) else x
return censored_text


There is an efficiency problem with your code. Python strings are immutable. Therefore, building strings using += concatenation is not optimal:

Concatenating immutable sequences always results in a new object. This means that building up a sequence by repeated concatenation will have a quadratic runtime cost in the total sequence length.

A way to avoid that performance problem would be to construct the result using ''.join(…) along with a generator expression:

def censor(text, word):
word_start = detect_word_start(text, word)
return ''.join(
'*' if word_start <= i < word_start + len(word) else x
for i, char in enumerate(text)
)


## Simple solution

There is a very simple solution, using str.replace() and the * operation on a string:

def censor(text, word):
return text.replace(word, '*' * len(word))


Let's start by removing print from the loop and add a return value

def censor(text,word):
#[...]
else:
censored_text += x
return censored_text


now we have a testable function. we add some tests

import unittest

class TestSome(unittest.TestCase):

def test_not_found(self):
self.assertEqual(censor("", "bar"), "")
self.assertEqual(censor("f", "bar"), "f")
self.assertEqual(censor("foo", "bar"), "foo")
self.assertEqual(censor("fooo", "bar"), "fooo")

def test_found(self):
self.assertEqual(censor("bar", "bar"), "***")
self.assertEqual(censor("bar!", "bar"), "***!")
self.assertEqual(censor("cow bar", "bar"), "cow ***")

def test_parts(self):
self.assertEqual(censor("foobar", "bar"), "foobar")
self.assertEqual(censor("bare", "bar"), "bare")

def test_capital(self):
self.assertEqual(censor("Bar", "bar"), "***")

def test_multiple(self):
self.assertEqual(censor("foo bar bar foo", "bar"), "foo *** *** foo")

if __name__ == '__main__':
unittest.main()


Running the tests show that the functions are not meeting my expectations.

• detect_word_start() does not return a value if nothing found
• you miss words at the beginning of a sentence becaue of capitalization
• you censor different words containing the character sequence
• you miss multiple appearences
• Perhaps it would be worthwhile to explain/link to the concept of TDD? IE, that it might have been easier to create the tests before the implementation and code from there. – Matthew FitzGerald-Chamberlain Aug 3 '18 at 17:50
• And then we run into the eternal question of "how much testing is enough?" Personally, I'd at least like to test that censoring bar will not change ba, babar or barbar, and that it will correctly censor BAR, (bar) and bar@bar.foo. Oh, and at least one test where the censored word is something other than bar! – Ilmari Karonen Aug 3 '18 at 18:19