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

AnnounceL: exam monday eve, no lecture monday

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
• Really you want experience with loops and complex logic
  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. As a practical matter, you would not go through 0..n-1 thais way, but it does illustrate the while loop structure to go through a series o numbers.

• 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
• Minor style point
  i < n is equivalent to i <= n + 1
  CS preference for the < form

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 at 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 approach for this problem. Here just showing for/while equivalence.

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

While Loop - Infinite Loop Danger

With the for/range loop, you never get an infinite loop. But with the while loop, it's not so hard to get an infinite loop — say you forget the i += 1 step, or write the test incorrectly. The while loop is more powerful, and predictably, it has more risk of screwups, so we use it only for cases where the simple for/range is insufficient.

Preface 1 - Advance With var += 1

• Framing for today's examples
• Imagine a string on paper
• Your finger is pointing at a char
• Move your finger to the right repeatedly, 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

Preface 2 - When is i In Bounds?

Suppose the int i is indexing into a string, and I am changing it with i += 1 or i -= 1. What are the bounds for i remaining a valid index into the string?

i < length

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.

i >= 0

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.

Python detail: surprisingly s[-1] does not give an error in Python, it accesses the last char. For our algorithms though, we treat i >= 0 as the boundary.

Aside - the CS106A Story Arc

• Have some problem IRL
• Think about what we want - a drawing or scheme about a data structure
• Think about steps on the drawing
• Translate into code
• Debug the code,
  Look at the error message
  Look at the drawing
• Add fixes to the code until it works
• Profit!

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]

Note This Works ok: 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 Tolerates 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.

Starter code

def exclamation(s):
    exclaim = s.find('!')
    if exclaim == -1:
        return ''

    # Your code here
    start = ???

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

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
  "precedence" in CS parlance
• Say have three boolean variables, each True/False
  age - say age is good if less than 30
  is_raining - True if raining
  is_weekend- True if it's the weekend
• Define: to be a good day, need two things:
  1. it must not be raining
  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 (and evaluates before or):

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

You can tell the above does not work right, because any time is_weekend is True, the whole thing is True, regardless of age or rain. 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 to set the order you want
• We will never complain about extra parenthesis, so add them to spell out 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')

(optional) Boolean Exercise oh_no()

(Got this far in lecture - exercise TBD)

> oh_no()

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]

Below here optional

Keyboard Accelerators

Command-return = run in exp server

Command-/ = comment, uncomment

Tab, Shift-tab = indent, unindent

Ctrl-k = delete line. Works in Gmail and in most browser forms. Super satisfying!

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

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 always name it "i" since that's the idiom programmers use for that context, so you cannot be blamed for thinking it was some Python rule.

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. The computer does not understand English here; it just recognizes that two words are the same and so must be the same variable.