How does list indexing work?

Question

This question is in python:

battleships = [['0','p','0','s'], ['0','p','0','s'], ['p','p','0','s'], ['0','0','0','0']] def fun(a,b,bships): c = len(bships) return bships[cb][a-1] print(fun(1,1,battleships)) print(fun(1,2,battleships)) 

first print gives 0 second print gives p

I cannot understand why, if you could give an explanation, it would be very helpful.

Thanks to those who help :)

Answer 1

Indexing starts at 0 . Thus, battleships contain elements with indices 0 , 1 , 2 , 3 .

The first len(bships) gets the length of the list of battleships lists, which is 4.

bships[cb][a-1] refers to the items in the list through their index value. Therefore, when you first call the function:

 print(fun(1,1,battleships)) 

This is bships[4-1][1-1] , which is bships[3][0] , which ['0','0','0','0'][0] , which 0

Answer 2

You can easily execute it by replacing the calculations with the actual values:

In the first call, you index:

 bships[cb][a-1] == bships[4-1][1-1] == bships[3][0] 

Count from 0 that the last line, ['0','0','0','0'] , the first element, '0' .

The second call is evaluated as:

 bships[cb][a-1] == bships[4-2][1-1] == bships[2][0] 

therefore, the first cell of the second row, ['p','p','0','s'] is 'p' .

Note that in Python you can use negative indices without first evaluating len() ; remove c from your function and everything will work the same:

 >>> battleships = [['0','p','0','s'], ... ['0','p','0','s'], ... ['p','p','0','s'], ... ['0','0','0','0']] >>> def fun(a,b,bships): ... return bships[-b][a-1] ... >>> print(fun(1,1,battleships)) 0 >>> print(fun(1,2,battleships)) p 

This is because Python processes negative indices from the end; internally, it will use the length of the sequence (which is stored in the sequence) to calculate the same thing, but faster.

Answer 3

 >>> battleships = [['0','p','0','s'], ... ['0','p','0','s'], ... ['p','p','0','s'], ... ['0','0','0','0']] >>> >>> a = 1 >>> b = 1 >>> c = len(battleships) >>> cb,a-1 (3, 0) 

Now battleships[cb][a-1] can be divided into two parts:

battleships[cb] and [a-1]

First, Python calls battleships[cb] as cb is 3 so that it returns the last list (4th element) from the ships of the line. ie ['0','0','0','0'] (indexing starts at 0 )

Now comes the second part: [a-1]

Now [a-1] is called into this returned list, i.e. ['0','0','0','0']

['0','0','0','0'][a-1] as a-1 is 0 , so python returns the first element from this list.

so you get "0".

The same applies for different values ​​of a , b :

 >>> a = 1 >>> b = 2 >>> cb,a-1 (2, 0) >>> battleships[cb] ['p', 'p', '0', 's'] >>> battleships[cb][a-1] #calls ['p', 'p', '0', 's'][0] 'p' 

Answer 4

If you have a problem understanding any new thing in programming, change the program a bit.

I will give you an example. I changed it a bit.

 battleships = [['0','p','0','s','3'], ['0','p','0','s','8'], ['p','p','0','s','2']] print "len(battleships) =",len(battleships) print "battleships[0] =",battleships[0] print "battleships[1] =",battleships[1] print "battleships[2] =",battleships[2] print "len(battleships[0]) =", len(battleships[0]) 

When I launched it, the output will be

 len(battleships) = 3 battleships[0] = ['0', 'p', '0', 's', '3'] battleships[1] = ['0', 'p', '0', 's', '8'] battleships[2] = ['p', 'p', '0', 's', '2'] len(battleships[0]) = 5 

Match output statements with print operations. This will help.

Try print battleships[0][1] etc.

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