Arrays (Lists) in Python

Arrays (Lists) in Python

one thing after another

Problem Given 5 numbers, read them in

and calculate their average THEN print out the ones that were

above average

Data Structure Needed Need some way to hold onto all the

individual data items after processing them

making individual identifiers x1, x2, x3,... is not practical or flexible

the answer is to use an ARRAY a data structure - bigger than an

individual variable or constant

An Array (a List) You need a way to have many variables

all with the same name but distinguishable!

In math they do it by subscripts or indexes x1, x2, x3 and so on

In programming languages, hard to use smaller fonts, so use a different syntax x [1], x[0], table[3], point[i]

Semantics numbered from 0 to n-1 where n is

the number of elements 0 1 2 3 4 5

Properties of an array (list) Heterogeneous (any data type!) Contiguous Have random access to any element Ordered (numbered from 0 to n-1) Number of elements can change very

easily (use method .append) Python lists are mutable sequences of

arbitrary objects

Syntax Use [] to give initial value to, like x =

[1,3,5] refer to individual elements

uses [ ] with index in the brackets most of the time you don’t refer to the

whole array as one thing, or just by the array name (one time you can is when passing a whole array to a function as an argument)

Python Programming, 2/e 8

List Operations you knowOperator Meaning

<seq> + <seq> Concatenation<seq> * <int-

expr>Repetition

<seq>[] Indexinglen(<seq>) Length<seq>[:] Slicing

for <var> in <seq>:

Iteration

<expr> in <seq> Membership (Boolean)

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Indexing an Array The index is also called the subscript In Python, the first array element always

has subscript 0, the second array element has subscript 1, etc.

Subscripts can be variables – they have to have integer values

k = 4 items = [3,9,’a’,True, 3.92] items[k] = 3.92 items[k-2] = items[2] = ‘a’


List Operations Lists are often built up one piece at

a time using append.nums = []x = float(input('Enter a number: '))while x >= 0: nums.append(x) x = float(input('Enter a number: '))

Here, nums is being used as an accumulator, starting out empty, and each time through the loop a new value is tacked on.


List OperationsMethod Meaning

<list>.append(x) Add element x to end of list.

<list>.sort() Sort (order) the list. A comparison function may be passed as a parameter.

<list>.reverse() Reverse the list.

<list>.index(x) Returns index of first occurrence of x.

<list>.insert(i, x) Insert x into list at index i.

<list>.count(x) Returns the number of occurrences of x in list.

<list>.remove(x) Deletes the first occurrence of x in list.

<list>.pop(i) Deletes the ith element of the list and returns its value.

Using a variable for the size It is very common to use a variable to

store the size of an array SIZE = 15 arr = [] for i in range(SIZE):

arr.append(i) Makes it easy to change if size of

array needs to be changed

Solution to starting problemSIZE = 5n = [0]*SIZEtotal = 0for ct in range(SIZE):

n[ct] = float(input("enter a number “))total = total + n[ct]

cont'd on next slide

Solution to problem - cont'd

average = total / SIZEfor ct in range(5):

if n[ct] > average: print (n[ct])

Scope of counter in a for loop The counter variable has usual

scope (body of the function it’s in) for i in range(5):

counter does exist after for loop finishes

what‘s its value after the loop?

Initialization of arrays a = [1, 2, 9, 10] # has 4 elements a = [0] * 5 # all are zero

Watch out index out of range! Subscripts range from 0 to n-1 Interpreter WILL tell you if an index

goes out of that range BUT the negative subscripts work as

they do with strings (which are, after all, arrays of characters)

x = [5]*5 x[-1] = 4 # x is [5,5,5,5,4]

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Assigning Values to Individual Array Elements

temps = [0.0] * 5m = 4temps[2] = 98.6;temps[3] = 101.2;temps[0] = 99.4;temps[m] = temps[3] / 2.0;temps[1] = temps[3] - 1.2; // What value is assigned?

temps[0] temps[1] temps[2] temps[3] temps[4]

7000 7004 7008 7012 7016

99.4 ? 98.6 101.2 50.6

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What values are assigned?SIZE =5temps = [0.0]* SIZE for m in range(SIZE):

temps[m] = 100.0 + m * 0.2

for m in range(SIZE-1, -1, -1):print(temps[m])


7000 7004 7008 7012 7016

? ? ? ? ?

Indexes Subscripts can be constants or

variables or expressions If i is 5, a[i-1] refers to a[4] and a[i*2]

refers to a[10] you can use i as a subscript at one

point in the program and j as a subscript for the same array later - only the value of the variable matters

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Variable Subscriptstemps = [0.0]*5m = 3 . . . . . .

What is temps[m + 1] ?

What is temps[m] + 1 ?


7000 7004 7008 7012 7016

100.0 100.2 100.4 100.6 100.8

Random access of elements Problem : read in numbers from a file,

only single digits - and count them - report how many of each there were

Use an array as a set of counters ctr [0] is how many zero's, ctr[1] is how

many ones, etc. ctr[num] +=1 is the crucial

statement

Parallel arrays Sometimes you have data of different

types that are associated with each other

like name (string) and GPA (float) You CAN store them in the same array

ar = [“John”, 3.24, “Mary”, 3.9, “Bob”, 2.7] You can also use two different arrays

"side by side"

Parallel arrays, cont'dfor i in range(SIZE): name[i], gpa[i] = float(input(“Enter”))Logically the name in position i corresponds to the gpa in position iNothing in the syntax forces this to be true, you just have to program it to be so.

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Parallel Arrays

Parallel arrays are two or more arrays that have the same index range and whose elements contain related information, possibly of different data types

EXAMPLE SIZE = 50 idNumber = [“ “]*SIZE hourlyWage = [0.0] *SIZE parallel

arrays

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SIZE = 50idNumber = [“ “] *SIZE // Parallel arrays holdhourlyWage =[0.0] *SIZE // Related information

idNumber[0] 4562 hourlyWage[0] 9.68



. . . . . . . . . . . .



Selection sort - 1-d arrayAlgorithm for the sort1. find the maximum in the list2. put it in the highest numbered element

by swapping it with the data that was at that location

3. repeat 1 and 2 for shorter unsorted list - not including highest numbered location

4. repeat 1-3 until list goes down to one

Find the maximum in the list# n is number of elementsmax = a[0] # value of largest element

# seen so farfor i in range(1, n): # note start at 1, not 0

if max < a[i]:max = a[i]

# now max is value of largest element in list

Find the location of the maxmax = 0 # max is now location of

the # largest seen so farfor i in range(1,n):

if a[max] < a[i]:max = i

# now max is location of the largest in # array

Swap with highest numberedRemember element at right end of

list is numbered n-1 temp = a[max] a[max] = a[n-1] a[n-1] = temp# there is a shorter way in Python!

The Python way! The previous code of finding the

max and its location will work in ANY high-level language.

Python has some nice functions and methods to make it easier!

Let’s try that.

The Python Way To find the max of the whole list mx = max(a) loc = a.index(mx) Is using index SAFE here? If it doesn’t

find mx in a, it will crash!But you just got mx from the list using

the max function, so it IS in the list a.

The Python Way The swap then becomes a[loc], a[n-1] = a[n-1],a[loc] Python “hides” the temporary third

variable

Find next largest element and swap (generic way)max = 0;for i in range(1,n-1): # note n-1, not n

if a[max] < a[i]:max = i

temp = a[max]a[max] = a[n-2] a[n-2] = temp

put a loop around the general code to repeat for n-1 passesfor pss in range(n, 1, -1): max = 0

for i in range(1,pss): if a[max] <= a[i]:max = i temp = a[max]

a[max] = a[pss-1] a[pss-1] = temp

The whole thing the Python way for pss in range(n, 1, -1): # n-1 passes mx = max(a[0:pss]) loc = a.index(mx)

a[loc], a[pss-1] = a[pss-1], a[loc]

2-dimensional arrays Data sometimes has more

structure to it than just "a list" It has rows and columns You use two subscripts to locate an

item The first subscript called “row”,

second called “column”

2-dimensional arrays syntax

a = [[0]*5 for i in range(4)] # 5 columns, 4 rows

Twenty elements, numbered from [0][0] to [4][3]

a = [[0]*COLS for i in range(ROWS)] Which has ROWS rows and COLS columns in

each row (use of variables to make it easy to change the size of the array without having to edit every line of the program)

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[0] [1] [2] [3] [4] [5] [6] [7] [8] [9] [10][11]

66 64 72 78 85 90 99 105 98 90 88 80row 2,col 7might beArizona’shigh forAugust

EXAMPLE -- Array for monthly high temperatures for all 50 states

NUM_STATES = 50NUM_MONTHS = 12stateHighs = [[0]*NUM_MONTHS for i in range(NUM_STATES)]

[0] [1] [2] . .

stateHighs[2][7] . [48] [49]

Processing a 2-d array by rowsfinding the total for the first row

for i in range(NUM_MONTHS):total = total + a[0][i]

finding the total for the second rowfor i in range(NUM_MONTHS):total = total + a[1][i]

Processing a 2-d array by rows

total for ALL elements by adding first row, then second row, etc.

for i in range(NUM_STATES):for j in range(NUM_MONTHS):

total = total + a[i][j]

Processing a 2-d array by columns

total for ALL elements by adding first column, second column, etc.

for j in range(NUM_MONTHS):for i in range(NUM_STATES):

total = total + a[i][j]

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Finding the average high temperature for Arizona

total = 0for month in range(NUM_MONTHS): total = total + stateHighs[2][month]average = round (total / NUM_MONTHS)

average

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Passing an array as an argument Arrays (lists) are passed by

reference = they CAN be changed permanently by the function

Definition def fun1 (arr): Call the function as

x = fun1 (myarr)

Arrays versus Files Arrays are usually smaller than files Arrays are faster than files Arrays are temporary, in RAM - files

are permanent on secondary storage Arrays can do random or sequential,

files we have seen are only sequential

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Using Multidimensional Arrays

Example of three-dimensional array

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NUM_DEPTS = 5 # mens, womens, childrens, electronics, furniture NUM_MONTHS = 12 NUM_STORES = 3 # White Marsh, Owings Mills, TowsonmonthlySales = [[[0]*NUM_MONTHS for i in range(NUM_DEPTS)]

for j in range(NUM_STORES)]

monthlySales[3][7][0]sales for electronics in August at White Marsh

12 MONTHS columns

5 D

EPTS

ro

ws

3 STORES

sheets

Example of filling a 3-d arraydef main(): NUM_DEPTS = 5 # mens, womens, childrens, electronics, furniture NUM_MONTHS = 12 NUM_STORES = 3 # White Marsh, Owings Mills, Towson monthlySales = [[[0]*NUM_MONTHS for i in range(NUM_DEPTS)] for j in range(NUM_STORES)] storeNames = ["White Marsh", "Owings Mills", "Towson"] deptNames = ["mens", "womens", "childrens", "electronics", "furniture"] for store in range(NUM_STORES): print (storeNames[store], end=" ") for dept in range(NUM_DEPTS): print (deptNames[dept], end = " ") for month in range(NUM_MONTHS): print("for month number ", month+1) monthlySales[store][dept] [month] = float(input("Enter the sales ")) print() print() print (monthlySales)

Find the average of monthly_salestotal = 0for m in range(NUM_MONTHS):

for d in range(NUM_DEPTS):for s in range(NUM_STORES):total += monthlySales[s][d][m]

average = total / (NUM_MONTHS * NUM_DEPTS *

NUM_STORES)

Problem: student data in a file The data is laid out as Name, section, gpa

John Smith, 15, 3.2 Ralph Johnson, 12, 3.9 Bob Brown, 9, 2.5 Etc.

Read in the datainf = open(“students”,”r”)studs = []for line in inf:

data = line.split(“,”)studs.append(data)

inf.close()#studs looks like [[“John Smith”,15,3.2], #[“Ralph Johnson”,12,3.9],[“Bob Brown”…]]

Find the student with highest GPA

max = 0for j in range(1, len(studs)):

if studs[max][2] < studs[j][2]:max = j

#max is now location of highest gpastuds[max][0] is the name of the studentstuds[max][1] is the student’s section

Arrays (Lists) in Python

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