Chapters 5&6: Iteration and Strings

Question 1

Increment/decrement

Answer 1

Updating the value of a variable bu increasing/drecreasing by 1 (or another fixed amount?) eg x=x+1

Question 2

while

Answer 2

statement saying “if this condition is met (ie True), execute the body. If False, skip rest of body and execute subsequent lines. Method of looping.

Question 3

n=10
while n>0:
   print(n)
   n=n-1
print('Blastoff!')
#number ofiterations?

Answer 3

10, 9, 8, 7, 6, 5, 4, 3, 2, 1, Blastoff!

10 interations

Question 4

iteration variable

Answer 4

variable which is changed each iteration and controls when the loop finishes, n above. If there isnt one, it’s an infinite loop.

Question 5

while True

Answer 5

A way of making a controlled infinite loop, which will repeat (if user input is part of it, ie. x=input(‘type Hi\n’)) until broken eg by user entering sought-for value, ie if x==’Hi’//[tab]break.

Question 6

while True:
   line=input('> ')
   if line[0]=='#'
      continue
   if line=='done'
      break
   print(line)
print('Done!')

Answer 6

line[0] means first character of value of line; hence if it starts with #, start loop again without finishing.
if it’s ‘done’, loop breaks and final line occurs. Else, loops back.

Question 7

Definite loop

Answer 7

A loop which goes through a set of values, and ends with the last value. Often with ‘for’ statement, rather than ‘while’ for an indefinite loop.

Question 8

friends=[‘Joseph’, ‘Glenn’, ‘Sally’]
for friend in friends:
print(‘Happy New Year: ‘, friend)
print(‘Done!’)

Answer 8

the ‘for’ makes it cycle through 3 times, on per friend in set ‘friends’, before executing final line. Friend is iteration variable.

Question 9

itervar

Answer 9

iteration variable, eg:
total=0
for itervar in [3, 41, 12, 9, 74, 15]:
   total=total+itervar
print('Total: ', total) #this would summate the numbers in set. (can just use sum() though)

Question 10

fruit=’banana’
letter=fruit[1]
print(letter)

Answer 10

Returns ‘a’. first letter is [0], b, etc. Must be integer

Question 11

len()

Answer 11

length of string, number of characters in a string (not other types)

Question 12

length=len(fruit)

last=fruit[length]

Answer 12

error - out of range! [6] would be 7th letter, only 6. hence last=fruit[length-1] to get last letter, a. OR, count back, letter=fruit[-1] returns a, [-2] b etc.

Question 13

spell out “fruit” letter by letter with “while” statement - traversing through a string

Answer 13

index=0

while index

Question 14

spell out “fruit” letter by letter with “for” statement - traversing through a string

Answer 14

for char in fruit:

print(char)

Question 15

String slices

Answer 15

assigning segments of a string for something, eg. s=’Ben Nouhan’, print(s[0:5]) returns Ben NB, it inludes first (0) but excludes last (5). [:5] is start to 5, [3:] is 3 to last. [:] is whole thing.

Question 16

Change assignmet of variable by 1 letter, eg Ben Nouhan to Ben Nouyan

Answer 16

s=’Ben Nouhan’

s=s[:7]+’y’+s[8:]

Question 17

count ‘a’s in banana

Answer 17

count=0
word='banana'
for letter in word:
   if letter=='a'
      count=count+1
print(count)

Question 18

letter in str

Answer 18

letter in a string. can be used with for, ie for letter in str:. an also be used as argument when defining funcction, ie def count(str, letter): etc.

Question 19

char in str

Answer 19

same as letter in str, but characters in general.

Question 20

in

Answer 20

A boolean operator, takes 2 strings and returns True if 1st appears as substring in 2nd, ie ‘a’ in ‘banana’ returns True.

Question 21

Alphabetical ordering

Answer 21

< and > can be used to compare string alphabeticaclly. ‘aaron’‘Jeff’. If you don’t want cacpitalised words to have priotity, need to cconvert all to lowercase.

Question 22

Object

Answer 22

eg a string. Contains data (eg the charaters of the string) and built-in methods, which are functions built into the given object, eg string-specific ones.

Question 23

dir()

Answer 23

used like type(), except rather than giving the type of objetc, it lists all the available methods avalable for that object type.

Question 24

help(object.method)

Answer 24

function for input of object type and method you want elucidating, eg help(str.capitalize) will explain that method for strings.

Question 25

method vs function syntax

Answer 25

variable.method() vs function(variable). Brackets contain parameters if applicacble. eg word.find(‘a’)

Question 26

Invocation

Answer 26

the call of a method. ie in above example, we invoked [method] on [variable], eg invoke find on word

Question 27

find method

Answer 27

word=banana
index=word.find(‘a’, 2)
print(index) #returns 3. a is the substring it looks for in word string, 2 is the letter (first n) it starts looking from. NB if non found, returns ‘-1’

Question 28

strip method

Answer 28

line=’ here we go ‘

line.strip() #returns ‘here we go’, removes spaces, tabs, enters from outside of non-white space.

Question 29

startswith method

Answer 29

line=’Ben Nouhan’

line.startswith(‘Ben’) #returns True, as it does. Case sensitive; b will return False, Be will return True.

Question 30

count method

Answer 30

s=banana

s. count(‘a’) #returns 3.
s. count(‘a’, 1, 5) #returns 2, as it counts ‘a’s between bAnaNa A and N inclusive

Question 31

help.method square brackets meaning

Answer 31

paramaters are listed by comma as usual. Square brackets indicate a parameter is optional. square brackets within eachother mean if you use the first parameter, the second is then optional.

Question 32

find unknown substr, z, between 2 known substrs, x&y in str

Answer 32

data=’str’
xpos=data.find(x)
ypos=data.find(y)
z=data[xpos+1:ypos]

Question 33

% (used on str)

Answer 33

format operator

Question 34

format sequences %d, %g, %s

Answer 34

integer, floating-point, string

Question 35

formatting example, eg with 3 format sequences

Answer 35

‘in %d years I have spotted %g %s.’%(3, 0.1, ‘camels’) #returns… well it’s obvious. number of parameters must match number of format sequences.