map,lambda,sorted

Aladdin

Oct 23, 2019 09:54

# name: 3_fn_lambda.py

# function with no name
# has a list of parameters
# :
# a single expression
lambda x : x * x
print("res : ", (lambda x : x * x)(10))
f1 = lambda x, y : x * y
f2 = lambda x, y : x * y

print(f1, type(f1))
print(f2, type(f2))

print(f1(10, 20))
print(f2(10, 20))

# sort
a = ['abcd', 'XyZ', 'Abef', 'Bx', 'zss']
print(sorted(a))
# sort in a case insensitive way
print(sorted(a, key = str.upper))

a = [ 'abcd', 'axae', 'xcyz', 'xwxz' ]
# combine 0th and 2nd char
print(sorted(a, key = lambda s : s[0] + s[2]))

John Rennie

Give me a moment and I'll copy and run the code.

Aladdin

i don't undertand how the first one works

John Rennie

Do you mean: lambda x : x * x ?

Aladdin

print("res : ", (lambda x : x * x)(10))

what is 10 doing there

how do you use lambda

i wrote a program but it's not working

John Rennie

The expression lambda x : x * x evaluates to a function. If you type it at the console you'll see it eavluates to:

>>> lambda x : x * x
<function <lambda> at 0x00000272212070D8>

Aladdin

Oct 23, 2019 10:02

yes memory locatin

i got it too

John Rennie

So if I type foo = lambda x : x * x this defines a new function foo() that squares its parameter.

Aladdin

how do you put value for x

i dont understand

John Rennie

>>> foo = lambda x : x * x
>>> foo(2)
4
>>> foo(3)
9

OK so far?

Aladdin

ok

John Rennie

But I can also skip assigning the new function to a variable like foo and just do:

>>> (lambda x : x * x)(2)
4
>>> (lambda x : x * x)(3)
9

Does that help?

Aladdin

Oct 23, 2019 10:05

yes

let me try

yep work

@JohnRennie What does sorted do

wait it is sorting

John Rennie

sorted is just a function that takes a list and returns a new list that is sorted.

Aladdin

so it's an extension of sort?

@JohnRennie a = ['abcd', 'XyZ', 'Abef', 'Bx', 'zss']
print(sorted(a))
# sort in a case insensitive way
print(sorted(a, key = str.upper))

John Rennie

mylist.sort does a sort in place i.e. it takes the list mylist and sorts it.

Aladdin

what is key

John Rennie

sorted(mylist) leaves mylist unchanged and returns a new list containing the same elements as mylist that is sorted.

So .sort() and sorted() both sort lists but they do it in different ways.

@Aladdin when you sort some collection you have to define how you want the sorting to work.

For example if you were sorting a box of apples you might choose to sort them by weight, or by size, or by how ripe they are and so on.

OK so far?

Aladdin

Oct 23, 2019 10:14

ok

John Rennie

In the sorted function the key parameter specifies how you want the list to be sorted.

Aladdin

ignore me

can u show example

John Rennie

An example of what?

Aladdin

sorted using key

John Rennie

We need to talk a bit about how sorting works.

Aladdin

Oct 23, 2019 10:18

ok

John Rennie

Suppose you have two strings x = "abc" and y = "def"

To sort the strings you need some way of finding out is x is less than, greater than or equal to y. OK so far?

Aladdin

ok

John Rennie

By default sorted compares the strings character by character. So it looks at x[0] and y[0] and sees which is greater, lesser or equal.

Aladdin

yes

John Rennie

If x[0] < y[0] then this means x < y and the comparison can stop immediately.

Aladdin

Oct 23, 2019 10:21

yep

John Rennie

If the first characters of the strings are identical then sorted moves on to look at x[1] and y[1].

If these are the same it moves on to x[2] and y[2] and so on until either it finds a difference or it comes to the end of the string.

Aladdin

yep

John Rennie

But you can tell sorted to use a different way of comparing the strings.

And that's what key is used for.

Suppose I use sorted(mylist, key=str.upper)

The value of key is a function that is applied to the elements before comparing them. So in this case sorted would start by comparing str.upper(x[0]) and str.upper(y[0]).

Aladdin

oh

John Rennie

So suppose x="aaa" and y="AAA"

if we compare the two we get y < x because the ASCII code for "A" is less than the ASCII code for "a". OK so far?

Aladdin

Oct 23, 2019 10:27

ok

John Rennie

But if we use key=str.upper then the function str.upper gets applied to the characters before sorted compares them. So in effect it is comparing str.upper(x) and str.upper(y) i.e. comparing "AAA" and "AAA". And of course they are the same.

Aladdin

yep

John Rennie

So by setting key=str.upper we can do a sort where it doesn't matter if letters are upper or lower case.

>>> a = ['abcd', 'XyZ', 'Abef', 'Bx', 'zss']
>>> sorted(a)
['Abef', 'Bx', 'XyZ', 'abcd', 'zss']
>>> sorted(a, key=str.upper)
['abcd', 'Abef', 'Bx', 'XyZ', 'zss']

Can you see how this works? With sorted(a) the uppercase letters come before the lowercase letters. With sorted(a,key=str.upper) the case doesn't matter.

Aladdin

yeah

John Rennie

This is a simple case, but the function you use as a key could be any function you want including functions you define yourself.

Aladdin

Oct 23, 2019 10:34

@JohnRennie Why str.upper

instead of a.upper?

John Rennie

Because str.upper is the name of a global function that uppercases its argument. a.upper is a method of a specific variable a.

>>> str.upper("abc")
'ABC'

Aladdin

ah

John Rennie

Actually I just checked and strings don't have a .upper method.

>>> a
['abcd', 'XyZ', 'Abef', 'Bx', 'zss']
>>> a.upper()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'list' object has no attribute 'upper'

Aladdin

aha thats why

@JohnRennie a = [ 'abcd', 'axae', 'xcyz', 'xwxz' ]

sorted(a,key=lambda a:a[0]+a[2])

why is this not working

John Rennie

>>> a = [ 'abcd', 'axae', 'xcyz', 'xwxz' ]
>>> sorted(a,key=lambda a:a[0]+a[2])
['axae', 'abcd', 'xwxz', 'xcyz']

It is working, but is it using a rather odd sort method.

Your sort function lambda a:a[0]+a[2] is adding together the first and third letters of the strings and using the result as the sort order.

Aladdin

Oct 23, 2019 10:49

it's not printing for me

John Rennie

You mean when you type sorted(a,key=lambda a:a[0]+a[2]) and press enter nothing happens?

Aladdin

no output is shown

no for this

a = [ 'abcd', 'axae', 'xcyz', 'xwxz' ]

sorted(a,key=lambda a:a[0]+a[2])

John Rennie

D:\rhs\Python>python
Python 3.7.4 (tags/v3.7.4:e09359112e, Jul  8 2019, 20:34:20) [MSC v.1916 64 bit (AMD64)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> a = [ 'abcd', 'axae', 'xcyz', 'xwxz' ]
>>>
>>> sorted(a,key=lambda a:a[0]+a[2])
['axae', 'abcd', 'xwxz', 'xcyz']

It works here. I just copied and pasted your code.

Aladdin

weird

John Rennie

Is that a blank line on your second line?

You haven't accidentally put some spaces there?

Aladdin

Oct 23, 2019 10:52

blank line

John Rennie

Don't know then.

Aladdin

@JohnRennie Sorry i put space

John Rennie

@Aladdin oops :-)

Aladdin

@JohnRennie a = [ 'abcd', 'axae', 'xcyz', 'xwxz' ]
# combine 0th and 2nd char
print(sorted(a, key = lambda s : s[0] + s[2]))

How does this work

John Rennie

The function you are specifying for the sorting is lambda s : s[0] + s[2]

Aladdin

Oct 23, 2019 10:58

why not a in there

John Rennie

This takes the first and third letters of the string and adds them.

@Aladdin it's the definition of a function

It's like:

def foo(s):
  return s[0] + s[2]

s is just the name the function uses for its parameter.

Aladdin

okay

@JohnRennie But it's not comining

combining 1st and 3nd character

John Rennie

s[0] is the first character of the string s and s[2] is the third character. And the function is returning s[0]+s[2].

>>> s="abc"
>>> s[0]+s[2]
'ac'

Aladdin

yes

but we want a not s

@JohnRennie hello?

print(sorted(a, key = (lambda s : s[0] + s[2]))(a))

John Rennie

s is just the name of the parameter as used in the function definition. It's like doing:

Aladdin

Oct 23, 2019 11:09

why so many brackets before a?

why not:

John Rennie

def foo(s):
  return s[0] + s[2]

a = "xyz"
foo(a)

Aladdin

print(sorted(a, key = (lambda s : s[0] + s[2])(a))

John Rennie

The variable you pass as the key to sorted has to be a function.

(lambda s : s[0] + s[2])(a) is not a function. It is the result returned by a function when you pass it the argument a.

Aladdin

ok

@JohnRennie How would change to a function

John Rennie

lambda s : s[0] + s[2] is the function.

Try typing that at the console and python will tell you it's a function:

>>> lambda s : s[0] + s[2]
<function <lambda> at 0x000001CF820D70D8>

Aladdin

Oct 23, 2019 11:16

so this code is wrong

a = [ 'abcd', 'axae', 'xcyz', 'xwxz' ]
# combine 0th and 2nd char
print(sorted(a, key = lambda s : s[0] + s[2]))

@JohnRennie

it can't combine oth and 2nd char

John Rennie

Lets simplify this to:

a = [ 'abcd', 'axae' ]
sorted(a, key = lambda s : s[0] + s[2])

So when we call sorted it has to compare the two strings 'abcd' and 'axae'. Yes?

Aladdin

yes

John Rennie

Without a key it would just compare the letters, but are telling it to use the function lambda s : s[0] + s[2] to do the comparison.

For the first string we get (lambda s : s[0] + s[2])('abcd') = 'ac'

For the second string we get (lambda s : s[0] + s[2])('axae') = 'aa'

And 'aa' < 'ae' so the second string is less than the first string.

Aladdin

how is s==a?

lik how is s taking values of a

John Rennie

We have a conceptual difficulty here ...

Aladdin

Oct 23, 2019 11:24

sorry it's not becoming clear to me

John Rennie

Are you happy that the key function is lambda s : s[0] + s[2] i.e. a function that takes a string as an argument and returns a new string made up from the first and third characters of the original string?

Aladdin

yep

John Rennie

And we'll use my simplified example with just two strings in the list:

a = [ 'abcd', 'axae' ]
sorted(a, key = lambda s : s[0] + s[2])

When you call sorted the sorted function takes its argument a and looks at the elements of the list a.

Aladdin

yes

John Rennie

In this case there are only two elements: 'abcd' and 'axae'

Aladdin

Oct 23, 2019 11:27

and then it applies the function on it

John Rennie

So the sorted function takes the two elements a[0] and a[1] and checks to see which is less.

And the way it does this is using the function lambda s : s[0] + s[2]

So the sorted function calls (lambda s : s[0] + s[2])(a[0]) and it also calls (lambda s : s[0] + s[2])(a[1]) and it looks at the reults returned by the function to see which is less.

Aladdin

ohkay

so argument is a

parmeter is s

and then sorts based on the result from the unction

John Rennie

Yes

Aladdin

@JohnRennie Lat doubt

John Rennie

OK ... ?

Aladdin

Oct 23, 2019 11:34

@JohnRennie can you tell me why set and tuple doesn't work

for map example

John Rennie

It's because when you call list(v) it empties the map object v and puts all its entries in your list.

>>> v=map(fun,a)
>>> v
<map object at 0x000001CF820DE0C8>
>>> list(v)
[1, 4, 9, 25]
>>> list(v)
[]

Aladdin

oh

John Rennie

So list(v) only works once. The second time it returns an empty list because the map object got emptied by the first call to list.

tuple and set will work if you call them immediately after the map

>>> v=map(fun,a)
>>> set(v)
{1, 4, 9, 25}

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