Today: parsing, while loop vs. for loop, parse words out of string patterns

Data and Parsing

Here's some fun looking data...

$GPGGA,005328.000,3726.1389,N,12210.2515,W,2,07,1.3,22.5,M,-25.7,M,2.0,0000*70
$GPGSA,M,3,09,23,07,16,30,03,27,,,,,,2.3,1.3,1.9*38
$GPRMC,005328.000,A,3726.1389,N,12210.2515,W,0.00,256.18,221217,,,D*78
$GPGGA,005329.000,3726.1389,N,12210.2515,W,2,07,1.3,22.5,M,-25.7,M,2.0,0000*71
$GPGSA,M,3,09,23,07,16,30,03,27,,,,,,2.3,1.3,1.9*38
$GPRMC,005329.000,A,3726.1389,N,12210.2515,W,0.00,256.18,221217,,,D*79
$GPGGA,005330.000,3726.1389,N,12210.2515,W,2,07,1.3,22.5,M,-25.7,M,3.0,0000*78
$GPGSA,M,3,09,23,07,16,30,03,27,,,,,,2.3,1.3,1.9*38
...

• The above is what a GPS chip outputs
• buried deep in your phone, this is going on
  It's a standard: NMEA_018
• Notice: it's just text
  a series of text lines ending with \n
  each line is made of chars
  We will use s.split(',') on lines like this later
• Text is a super common exchange format between systems
• "Parsing"
  Have raw text like this example
  Find and pull out the data you want
• On the surface, we are doing parsing examples today
• In reality, learning index/loop algorithms, parsing is just a handy domain

Recall: for i/range

The for/i/range form is great for going through numbers which you know ahead of time - a common pattern in real programs. If you need to go through 0..n-1 - use for/i/range, that's exactly what it's for. For example, if we want to loop over 0..5, say to index into 'Python'

Flexible loop: while

But we also have the while loop. The "for" is suited for the case where you know the numbers ahead of time. The while is more flexible. The while can test on each iteration, stop at the right spot. Ultimately you need both forms, but here we will switch to using while.

while Equivalent of for/range

It's possible to write the equivalent of for/i/range as a while loop instead. This is not a good way to go through 0..n-1, but it does show a way to structure a while loop.

• for i in range(n) - go-to solution for that sequence
• Can write this as a while .. do steps manually
• Three parts: init, test, update
• Use range() for common 0..n-1 case
• Use while where need fine control of i (examples to follow)
• Beware: easy to forget update step, result is infinite loop
  for/range is so common .. we don't have muscle-memory for the update line

Here is the while-equivalent to for i in range(n)

i = 0         # 1. init
while i < n:  # 2. test
    # use i
    i += 1    # 3. update, loop-bottom
              # (easy to forget this line)

Example while_double()

> while_double() (in parse1 section)

double_char() written as a while. The for-loop is the better appraoch approach for this problem, but using while here to show for/while equivalence.

def while_double(s):
    result = ''
    i = 0
    while i < len(s):
        result += s[i] + s[i]
        i += 1
    return result

Foreshadow i Valid:
i < length and i >= 0

With zero based indexing, if we are increasing an index variable i, then
i < length is the easy test that i is a valid index; that it is not too big.

If we are increasing an index number, 5 is the last valid index. When we increase it to 6 it's past the end of the string. The length here is 6, so in effect i < 6 checks that i is valid if we are increasing i.

If we are decreasing i, then i >= 0 is the valid check, since 0 is the first index. Could equivalently write this as i > -1, but usually it's written as i >= 0.

Foreshadow Advance With var += 1

• Framing for today's example
• Imagine a string on paper
• Finger is pointing at a char
• Move finger to the right, looking for something
• Python: have a var end storing an index into string
• e.g. end is 4, pointing at the 'a'
• end += 1 .. like moving one to the right
• Continue end += 1 until get to a space

Start with end = 4. Advance to space char with end += 1 in loop

Observe - the CS106A Lifestyle Pattern

• Have some problem IRL
• Think about what problem looks like in the computer - detail oriented
• Think about what we want
• Work out an algorithm to get there
• Then express it as code
• Then it doesn't work right the first 99 times
• Then it works, and we declare it perfect!
• Try this out on the at_word() problem below

Example: at_word()

> at_word() (in parse1 section)

'xx @abcd xyz' -> 'abcd'
'x@ab^xyz' -> 'ab'

at_word(s): We'll say an at-word is an '@' followed by zero or more alphabetic chars. Find and return the alphabetic part of the first at-word in s, or the empty string if there is none. So 'xx @abc xyz' returns 'abc'.

• Realistic parsing problem, extracting the wanted part of a string
• Demonstrate several patterns on this one
• We'll re-use these patterns
• We'll work through this one carefully
• Points about this code:
  Use str.find() to locate each @
  Use while to skip over alpha chars to find end
  Use var < len(s) to protect use of s[var]
  Var names: search, at, end - try to keep things straight

at_word() Strategy 1

First use s.find() to locate the '@'. Then start end pointing to the right of the '@'.

at_word() Start Picture

Code to set this up:

    at = s.find('@')
    if at == -1:
        return ''
    
    end = at + 1

at_word() Goal Picture

at_word() While Test

Use a while loop to advance end over the alphabetic chars. What is the test for this loop? Work it out on the drawing.

    while ???? 
        end += 1

• AKA skip over the alpha chars
• Loop test: this is true = advance end by one
• Test: s[end].isalpha()
• Reminisce : Bit "true = go" pattern for moving forward
• This code will 90% work, with one case to fix later
• Loop leaves end pointing to the first non-alpha char

This loop is 90% correct to advance end:

    # Advance end over alpha chars
    while s[end].isalpha():
        end += 1

at_word() Slice with end

Once we have at/end computed, pulling out the result word is just a slice.

    word = s[at + 1:end]
    return word

at_word() V1

Put those phrases together and it's an excellent first try, and it 90% works. Run it.

def at_word(s):
    at = s.find('@')
    if at == -1:
        return ''
    
    end = at + 1
    # Advance end over alpha chars
    while s[end].isalpha():
        end += 1

    word = s[at + 1:end]
    return word

at_word: 'woot' Bug

That code is pretty good, but there is actually a bug in the while-loop. It has to do with particular form of input case below, where the alphabetic chars go right up to the end of the string. Think about how the loop works when advancing "end" for the case below.

    at = s.find('@')
    end = at + 1
    while s[end].isalpha():
        end += 1

'xx@woot'
 01234567

Problem: keep advancing "end" .. past the end of the string, eventually end is 7. Then the while-test s[end].isalpha() throws an error since index 7 is past the end of the string.

The loop above translates to: "advance end so long as s[end] is alphabetic"

To fix the bug, we modify the test to: "advance end so long as end is valid and s[end] alphabetic".

In other words, stop advancing if end reaches the end of the string.

Loop end bug:

Solution: end < len(s) Guard Test

This "guard" pattern will be a standard part of looping over something. We cannot access s[end] when end is too big. Add a "guard" test end < len(s) before the s[end]. This stops the loop when end gets to 7. The slice then works as before. This code is correct.

def at_word(s):
    at = s.find('@')
    if at == -1:
        return ''

    # Advance end over alpha chars
    end = at + 1
    while end < len(s) and s[end].isalpha():
        end += 1
    
    word = s[at + 1:end]
    return word

Guard / Short Circuit Pattern

The "and" evaluates left to right. As soon as it sees a False it stops. In this way the < len(s) guard checks that "end" is a valid number, before s[end] tries to use it. This a standard pattern: the index-is-valid guard is first, then "and", then s[end] that uses the index. We'll see more examples of this guard pattern.

Fix End Bug Recap

• Bug: run end off the end of s, testing non-existent s[end]
• e.g. this happens if input is s = 'xx @woot'
  Think through how the loop works for that case
• Solution:
• Add guard <
• This is the fixed loop:
  while end < len(s) and s[end].isalpha():
• Q: How to test if index i is valid in s?
• A: i < len(s)
• Only look at s[end] char after checking that end is valid
• Boolean Short Circuit
  Python evaluates expression left-right
  As soon as boolean value determined, stops trying
  A False in the midst of an and stops
  So the < guards the s[end].isalpha()
• Common guard pattern:
  Check i < len(s) before trying s[i]

s[end] vs. s[at + 1:end]

• Accessing s[end] off the end of the string is an error
  So we need the guard
• What about the slice s[at + 1:end]
• That actually works fine, for two reasons...

Reason 1 - UBNI

• Why does s[at + 1:end] work fine?
• Up To But Not Including - UBNI
• The char at the second index is not included in the slice
• So the fact that it's one past the end is fine - it is not included
• This is the best reason, so focus on this one

Reason 2 - Slice Garbage

• The other reason it works is a little sketchy
• It turns out, slices never raise an error about bad out of bounds index numbers
• They will work with any old garbage numbers
• If a number is too big, it is interpreted as "the end of the string"
• This does not mean you can stop caring about index numbers
• It just means the slice is not checking for you
• Our above solution is fine - the end index is managed accurately
  Going exactly one past the chars we want in all cases

>>> s = 'Python'
>>> len(s)
6
>>> s[2:5]
'tho'
>>> s[2:6]
'thon'
>>> s[2:46789]
'thon'

at_words() - Zero Char Case - Works?

• What about 'xx @ xx'
• Consider slice of @ above
  s[at + 1:end]
  Turns out to be like s[4:4]
  Which is the empty string '', so the code we have works perfectly for this edge case

Example/Exercise: exclamation()

> exclamation()

exclamation(s): We'll say an exclamation is zero or more alphabetic chars ending with a '!'. Find and return the first exclamation in s, or the empty string if there is none. So 'xx hi! xx' returns 'hi!'. (Like at_word, but right-to-left).

Set a variable start to the left of the exclamation mark. Move it left over the alphabetic chars.

Will need a guard here, as the loop goes right-to-left. The leftmost valid index is 0, so that will figure in the guard test.

exclamation() Solution

def exclamation(s):
    exclaim = s.find('!')
    if exclaim == -1:
        return ''
        
    # Your code here
    # Move start left over alpha chars
    # guard: start >= 0
    start = exclaim - 1
    while start >= 0 and s[start].isalpha():
        start -= 1
    
    # start is on the first *non* alpha
    word = s[start + 1:exclaim + 1]
    return word

Recall: Boolean Precedence

• If there is a mixture of and with or
• and has higher precedence, so it is evaluated first
• In all cases, you can add parenthesis to indicate the order you want

Parse "or" Example - at_word99()

> at_word99()

'xx @ab12 xyz' -> 'ab12'

at_word99(): Like at-word, but with digits added. We'll say an at-word is an '@' followed by zero or more alphabetic or digit chars. Find and return the alpha-digit part of the first at-word in s, or the empty string if there is none. So 'xx @ab12 xyz' returns 'ab12'.

We've reached a very realistic level of complexity for solving real problems.

"end" Loop For at_words99()

Like before, but now a word is made of alpha or digit - many real problems will need this sort of code. This may be our most complicated line of code thus far in the quarter! Fortunately, it's a re-usable pattern for any of these "find end of xxx chars" problems.

The most difficult part is the "end" loop to locate where the word ends. What is the while test here? (Bring up at_word99() in other window to work it out). We want to use "or" to allow alpha or digit.

at = s.find('@')
end = at + 1
while ??????????:
    end += 1

at_word99() While Test

# 1. Still have the < guard
# 2. Use "or" to allow isalpha() or isdigit()
# 3. Need to add parens, since this has and+or
#    combination
while end < len(s) and (s[end].isalpha() or s[end].isdigit()):
    end += 1

at_word99() Solution

def at_word99(s):
    at = s.find('@')
    if at == -1:
        return ''

    # Advance end over alpha or digit chars
    # use "or" + parens
    end = at + 1
    while end < len(s) and (s[end].isalpha() or s[end].isdigit()):
        end += 1
    
    word = s[at + 1:end]
    return word

If we have time, we'll look at these.

(optional) Exercise: dotcom2()

> dotcom2()

 'xx www.foo.com xx' -> 'www.foo.com'

dotcomt2(s): We are looking for the name of an internet host within a string. Find the '.com' in s. Find the series of alphabetic chars or periods before the '.com' with a while loop and return the whole hostname, so 'xx www.foo.com xx' returns 'www.foo.com'. Return the empty string if there is no '.com'. This version has the added complexity of the periods.

Ideas: find the '.com', loop left-right to find the chars before it. Loop over both alphabetic and '.'

dotcom2() Solution

def dotcom2(s):
    com = s.find('.com')
    if com == -1:
        return ''
    
    # "or" logic - move leftwards over
    # alphabetic or '.'
    start = com - 1
    while start >= 0 and (s[start].isalpha() or s[start] == '.'):
        start -= 1
    
    return s[start + 1:com + 4]

Style: Long Lines

Normally each Python line of code is un-broken. BUT if you add parenthesis, Python allows the code to span multiple lines until the closing parenthesis. Indent the later lines an extra 4 spaces - in this way, they have a different indentation than the body of the while. There's also a preference to end each line with an operator like or .. to suggest that there's more on the later lines.

    while (end < len(s) and 
            (s[end].isalpha() or
            s[end].isdigit())):
        end += 1