Slide 1

Today: string in, string upper/lower, if/elif, string .find(), slices

Some of today's material is review.

See chapters in Guide:

• String
• If

Slide 2

Announcements

1. Assignment 3, Sand! has been released, and is due next Tuesday.
2. The first quiz is next Monday. It will cover Bit problems and image problems. The best study materials are:
   • The expermental server Bit and Image problems
   • Section 1 and 2 problems

Slide 3

String in Test

• String in - test if a substring appears in a string
• Chars must match exactly - upper/lower are considered different
• This is True/False test - use .find() (below) to know where substring is
• Mnemonic: Python re-uses the word "in" from the for loop
• Style: don't write code for something that Python has built-in

>>> 'c' in 'abcd'
True
>>> 'bc' in 'abcd'  # works for multiple chars
True
>>> 'bx' in 'abcd'
False
>>> 'A' in 'abcd'   # upper/lower are different
False

Slide 4

Strategy: Built In Functions

Python has many built-in functions, and we will see all the important ones in CS106A. You want to know the common built-in functions, since using a built-in is far preferable to writing code for it yourself - "in" is a nice example. The "in" test for several data structure to see if a value is in there. This is our first example with strings.

Slide 5

Example: has_pi()

has_pi(s): Given a string s, return True if it contains the substrings '3' and '14' somewhere within it, but not necessarily together. Use "in".

> has_pi()

Note these functions are in the string-3 section on the Experimental server

Slide 6

boolean-test AND boolean-test

This form works in English but not in Python:

if '3' and '14' in s:   # NO does not work
    ... 

The and should connect two fully formed boolean tests, such as you would write with "in" or "==", so this works

if '3' in s and '14' in s:
    ...

This problem has a one-liner solution that doesn't even use if:

return '3' in s and '14' in s

Slide 7

Practice: has_first()

> has_first()

Slide 8

Recall: String += Accumulate Pattern

• result = '' - at start
• result += xxxxx - in loop
• return result - at end

Slide 9

String Character Classes

• Strings are made of characters, chars
• Chars are divided into "classes"
• "alpha" - alphabetic "word" chars, e.g. a-z
  Other non-roman alphabets, have their own alphabetic chars
• "digit"" - 0-9
• "space" - space, tab, newline
• Python char test functions
  Work on 1 or more chars
  Returns True if true for all chars
  False for the empty string
• s.isdigit() - is digit char
• s.isalpha() - is alphabetic char
• s.isspace() - is space char
• Space is mentioned for completeness, we'll concentrate on alpha and digit

>>> 'a'.isalpha()
True
>>> 'cat'.isalpha()
True
>>> '5'.isalpha()
False
>>> '5'.isdigit()
True
>>> '@'.isalpha()
False

Slide 10

Uppercase / Lowercase chars

• In some languages, alpha chars can have upper/lower pairings
  'a' is the lowercase form of 'A'
'A' is the uppercase form of 'a'
• s.upper() - returns uppercase form of s
• s.lower() - returns lowercase form of s
• Immutable: s.upper() returns a new, converted string
  The original string s is unchanged
• s.isupper() - True if made of uppercase chars
• s.islower() - True if made of lowercase chars
• A char with no upper/lower difference, e.g. '@' or '2'
  Not changed by upper()/lower()
  isupper()/islower() return False

>>> 'Kitten123'.upper()  # return with all chars in upper form
'KITTEN123'
>>> 'Kitten123'.lower()
'kitten123'
>>> 
>>> 'a'.islower()
True
>>> 'A'.islower()
False
>>> 'A'.isupper()
True
>>> '@'.islower()
False
>>> 'a'.upper()
'A'
>>> 'A'.upper()
'A'
>>> '@'.upper()
'@'
>>> s = 'Hello'
>>> s.upper()    # Returns uppercase form of s
'HELLO'
>>>
>>> s            # Original s unchanged
'Hello'
>>>

Slide 11

Example: alpha_up()

> alpha_up()

'12abc34' -> 'ABC'

Given string s. Return a string made of all the alphabetic chars in s, converted to uppercase form.

Use string functions .isalpha() and .upper()

Solution

def alpha_up(s):
    result = ''
    for i in range(len(s)):
        if s[i].isalpha():
            result += s[i].upper()
    return result

Slide 12

Example: catty()

> catty()

'xCtxxxAax' -> 'CtAa'

Return a string made of the chars from the original string, whenever the chars are one of 'c' 'a' 't', (either lower or upper case). So the string 'xaCxxxTx' returns 'aCT'. (Had an earlier version of this function that was case-sensitive.)

Slide 13

Catty Version That Doesn't Work - V1

Here is a natural way to think of the code, but it does not work:

def catty(s):
    result = ''
    for i in range(len(s)):
        if s[i] == 'c' or s[i] == 'a' or s[i] == 't':
            result += s[i]
    return result
    

What is the problem? Upper vs. lower case. We are not getting any uppercase chars 'C' for example.

Slide 14

Catty Solution V2

Solution: convert each char to lowercase form .lower(), then test.

Solution - this works, but that if-test is ugly

def catty(s):
    result = ''
    for i in range(len(s)):
        if s[i].lower() == 'c' or s[i].lower() == 'a' or s[i].lower() == 't':
            result += s[i]
    return result

Slide 15

Idea: Decomp By Var

• The code is getting a little lengthy
• The repeated s[i].lower() is irksome
• Introduce a variable to hold a commonly used value
• Advantages
• 1. Shorten the code, less repetitive typing
• 2. Variable name helps the code "read" better
• 3. A sort of decomp within a function - break the big thing into little steps

Slide 16

Decomp Var Steps

• Some phrase X repeated in code
  e.g. s[i].lower()
• Used several times, or it's just wordy to type
• Create a variable, compute once and store
  low = s[i].lower()
• Use that variable on later lines
• Variable name noun - code "reads" better
• Aside: can name the var anything, and the code still works

Slide 17

Catty Solution V3 - Better

Create variable to hold the lengthy computation - shorter code and "reads" better.

    low = s[i].lower()

The complete solution

def catty(s):
    result = ''
    for i in range(len(s)):
        low = s[i].lower()   # decomp by var
        if low == 'c' or low == 'a' or low == 't':
            result += s[i]
    return result

Style aside: the name of a variable should remind us of the role of that data in the local code. Other than that, the name can be short. The name does not need to repeat every true thing about the value. Just enough to distinguish it from other values in this algorithm.

Good names, short but including essential details: low, low_char

Names with more detail, probably too long: low_char_i, low_char_in_s

The V2 code above is acceptable, but V3 is shorter and nicer. The V3 code also runs faster, as it does not compute the lowercase form three times per char.

Slide 18

if/elif Structure

• An extra "if" feature, not used so often
• if/elif - a series of if-tests
• Evaluate test1, test2, test3
• As soon as test is true
  Runs that action
  Exits the if/elif structure
  No further tests/actions will run
• Optional "else:" at end runs if no test is true
• The plain if-statement is used most often
• Only use this form when you have a series of tests
• Mnemonic: "else" and "elif" are the same length

if test1:
  action-1
elif test2:
  action-2
else:
  action-3

Slide 19

Example: str_adx()

> str_adx()

str_adx(s): Given string s. Return a string of the same length. For every alphabetic char in s, the result has an 'a', for every digit a 'd', and for every other type of char the result has an 'x'. So 'Hi4!x3' returns 'aadxad'. Use an if/elif structure.

• If/elif logic to check for different char types
• alpha char → 'a', digit → 'd', otherwise → 'x'
• e.g. 'Z5$$t' → 'adxxa'

Solution

def str_adx(s):
    result = ''
    for i in range(len(s)):
        if s[i].isalpha():
            result += 'a'
        elif s[i].isdigit():
            result += 'd'
        else:
            result += 'x'
    return result

Slide 20

String find()

• s.find(target_str) – search s for target_str
• Returns int index where found first, searching from start of s
• Returns -1 if not found Important!
• Alternate form: 2nd "start_index" parameter, starts search from there
  s.find(target_str, start_index) (use this later)

>>> s = 'Python'
>>> s.find('t')
2
>>> s.find('th')
2
>>> s.find('n')
5
>>> s.find('x')
-1
>>> s.find('N')
-1

Slide 21

Strategy: Dense = Slow Down

• Some lines of code are routine
• Require just normal attention
• An advantage of using idiomatic phrases
  for i in range(len(s)):
• But some lines are dense
• Slow down for those, work carefuly
• Slices (below) are dense!
• Dense = Powerful!

Slide 22

Python String Slices

• This is a fantastic feature
• "substring" - contiguous sub-part of a string
• Access substring with 2 numbers
• "slice" uses colon to indicate a range of indexes
• s[1:3] returns 'yt'
• Start at first number
• Up to but not including second number UBNI
• s[3:3] = empty string
  "Not including" dominates the "starting at"
• Try it in the interpreter
• Style: typically written with no spaces around ":"

>>> s = 'Python'
>>> s[1:3]    # 1 .. UBNI
'yt'
>>> s[1:5]
'ytho'
>>> s[4:5]
'o'
>>> s[4:4]    # "not including" dominates
''

Slide 23

Omit Start/End Index

• If start index is omitted, goes from start of string
• If end index is omitted, goes through end of string
• If number is too big .. uses end also
• Note perfect split: s[:4] and s[4:]
  Number used as both start and end
  Splits the string into 2 pieces exactly

>>> s[:3]     # omit = from/to end
'Pyt'
>>> s[4:]
'on'
>>> s[4:999]  # too big = through the end
'on'
>>> s[:4]     # "perfect split" on 4
'Pyth'
>>> s[4:]
'on'
>>> s[:]      # the whole thing
'Python'

Slide 24

Example: brackets()

A first venture into using index numbers and slices. Many problems work in this domain - e.g. extracting all the hashtags from your text messages.

'cat[dog]bird' -> 'dog'

> brackets

• Problem spec: either 2 brackets, or zero brackets
• Strategy:
• Use s.find()
  left = s.find('[')
right = s.find(']')
• Switch between drawing and code
• Decomp by var
  Store in variable left for later lines
  Nice to have words left and right in code narrative
• Look for right bracket
• Use slice to pull out and return answer

Slide 25

Brackets Drawing

Slide 26

Brackets Observations

• Make a drawing - work out the index numbers
• Diagram / example strategy
• Code should work in general
• BUT can use specific string to work out numbers
• e.g. 'cat[dog]bird'
• Empty string input - works?
• What about input 'a[]z'
  Verify that our slice works here too

Solution

def brackets(s):
    left = s.find('[')
    if left == -1:
        return ''
    right = s.find(']')
    # Use slice to pull out chars between left/right
    # make a drawing!
    return s[left + 1: right]

Slide 27

Brackets - Decomp By Var

The variables left and right make this code more readable. They work naturally in the drawing and in the code, naming an important intermediate value that runs through the computation.

Below is what the code looks like without the variable. It works fine, and it's one line shorter, but the readability is a worse. It also likely runs a little slower, as it computes the left-bracket index twice.

def brackets(s):
    if s.find('[') == -1:
        return ''
    return s[s.find('[') + 1: s.find(']')]

Our brackets solution with its variables looks better.

Slide 28

Aside: Off By One Error

Int indexing into something is extremely common in computer code. So of course doing it slightly wrong is very common as well. So common, there is a phrase for it - "off by one error" or OBO — it even has its own wikipedia page. You can feel some kinship with other programmers each time you stumble on one of these.

"My code is perfect! Why is this not working? Why is this not work ... oh, off by one error. We meet again!"

Slide 29

Practice: at_3()

Here is a problem similar to brackets for you to try. If we have enough time in lecture, we'll do it in lecture. A drawing really helps the OBO on this one.

> at_3

Slide 30

Optional: Negative Slice

• We'll cover this someday maybe
• Optional / advanced shorthand
• Handy to refer to chars at end of string instead of beginning
• Negative numbers to refer to chars at end of string
• -1 is the last char
• -2 is the next to last char
• Works in slices etc.
• Maybe just memorize this one:
  s[-1] is the last char in s

>>> s = 'Python'
>>> s[len(s)-1]
'n'
>>> s[-1]  # -1 is the last char
'n'
>>> s[-2]
'o'
>>> s[-3]
'h'
>>> s[1:-3]  # works in slices too
'yt'
>>> s[-3:]
'hon'
>>> s[:-1] # return all but the last characer (used very often!)