Lecture 13: String Parsing and Code Style

July 13th, 2021

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

Reminders

• HW3 is due tonight at 11:55pm PT
• Quiz results will be available after class today
• Always use == or != instead of is or is not
• Pycharm will try to correct you, but ignore that!

BlueScreen Contest

• View Submissions
• Link to Voting Form
• Vote by 1:50pm PT today

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
• A standard: NMEA
  - NMEA_018 (wikipedia)
• Things to notice: it's just text
  -a series of text lines ending with \n
  -each line is made of chars
• Text is a super common exchange format between systems
• "Parsing"
  -Have "raw" text like this
  -Find and pull out the data you want

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 i with an index into string
• e.g. i is 4, pointing at the 'a'
• i += 1 .. like moving one to the right
• e.g. looking for the space after 'abc'

'xx @abc x'
 012345678

Compare for i/range vs. while

The for/i/range form is great for going through numbers which you know ahead of time - a common pattern in real programs. However, while is more flexible - can test on each iteration, stop at the right spot. Ultimately you need both forms.

while Equivalent of for/range

Use for/i/range if have a series of numbers to step through. That is a common case, and for/i/range is perfect for it. We'll use while for situations that require more flexibility.

• 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 cases
• 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

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)

Example while_double()

> while_double() (in parse1 section)

double_char() written as a while (using a range() is the better approach for this problem, so here just demonstrating for-while equivalence).

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

Test is i < len(s) - this is basically "is i valid". We'l use this definition of valid index later.

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Example: at_word()

> at_word() (in parse1 section)

'xx @abc xyz' -> 'abc'

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 a while to skip over alpha chars to find end
  Use < len(s) to protect use of s[xxx]
  Var names: search, at, end - try to keep things straight

at_word() Strategy

First use s.find() to locate the '@'. Then start end pointing to the right of the '@'. Use a while loop to advance end over the alphabetic chars. Make a drawing below to sketch out this strategy. Think about the while-test.

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

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

'xx @abcd xx'

at_word() Loop: Advance End Over Alpha Chars

• AKA skip over the alpha chars
• Loop test: this is true = advance end one more
• 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

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

Start of loop:

End of loop:

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

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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 s[end] is valid and alphabetic".

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

Loop end bug:

Solution: end < len(s) Guard Test

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 check s[end] 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]

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 it works

Example 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).

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.

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Boolean Operators

• Arithmetic operators: + - * /
• Combine numbers to make a number
  1 + 2 * 3 -> 7
• Boolean operators: and or not
• Combine Boolean values to make a Boolean
  True or False -> True
True and False -> False
True and Not False -> True

Boolean Expressions

See the guide for details Boolean Expression

• Boolean operators: and or not
• Mixture of these, can add parenthesis to force order of operation
• Say have these variables:
  age - int age, say age is good if less than 30
  is_raining - boolean, True if raining
  is_weekend- boolean, True if it's the weekend
• Define: to be a good day, need two things:
  1. it must not be raining. This is mandatory.
  2. then either age is under 30 or it's the weekend

The code below looks reasonable, but doesn't quite work right

def good_day(age, is_weekend, is_raining):
    if not is_raining and age < 30 or is_weekend:
        print('good day')

Boolean Precedence:

• not = highest, (like - in -7)
• and = next highest (like *)
• or = lowest (like +)

What The Above Does

Because and is higher precedence than or as written above, the code above acts like the following (the and going before the or):

   if (not is_raining and age < 30) or is_weekend:

What is a set of data that this code will evaluate incorrectly? raining=True, age=anything, weekend=True .. the or weekend makes the whole thing True, no matter what the other values are. This does not match the good-day definition above, which requires that it not be raining.

Boolean Precedence Solution

The solution we will spell out is not difficult.

• Many programmers do not have boolean precedence memorized .. fine
• Do remember that "not" is the highest precedence
• Solution: note when you have a mixture of and + or
  When there is a mixture, the precedence will matter
  put in parenthesis in that case
• We will never complain about parenthesis, so just add them to force the order you want
• In this case, put parens to group the or part, separating from not-raining
• BTW similar logic applies to math - if there's a mixture of * and +, add parenthesis

Solution

def good_day(age, is_weekend, is_raining):
    if not is_raining and (age < 30 or is_weekend):
        print('good day')

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Parse "or" Example - at_word99()

This is operating at a realistic level for parsing data.

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'.

"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 xxxx" 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

'@a12'
 01234

Solution

# 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

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We Need To Have a Little Talk About Variable Names

With the following code, it's clear that the assignment = sets the variable to point to a value.

x = 7

For Loop Sets Variables Too

It's less obvious, but the for loop just sets a variable too, once for each iteration. The variable name is the word the programmer chooses right after the word "for", in this example the variable is i which is an idiomatic choice:

for i in range(4):
    # use i
    print(i)

0
1
2
3

Variables and Meaninglessness

The Sartre of Coding!

The variable name is just the label applied to the box that hold the pointer.

You might get the feeling in CS106A to this point: it will only work if the variable is named "i", but that's not true. We named it "i" since that's the idiom, but it's not a requirement.

We try to choose a sensible label to keep our own thoughts organized. However the computer does not care about the word used, so long as the word chosen is used consistently across lines. The variable name i is idiomatic for that sort of loop. But in reality we could use any variable name, and the code would work exactly the same. Say we name the variable meh instead .. same output. All that matters is that the variable on line 1 is the same as on line 2.

for meh in range(4):
    print(meh)

0
1
2
3

This is a little disturbing. We do try to choose good and/or idiomatic variable names for our own sake. However, the computer does not notice or care about the actual word choice for our variables.