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Bad index in a sublist

asked 2021-03-21 11:27:15 +0200

Cyrille gravatar image

In the following code

A=[[99999,99999,1,99999],[1,99999,1,99999]]
A[0].index(A[0][2])
A[0].index(A[0][0])
A[0].index(A[0][1])

The index returned is the true one if the value is encountered form the first time in the 0th element of A. So A[0].index(A[0][2]) and A[0].index(A[0][0]) give the good index. But not A[0].index(A[0][1]). Why ?

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answered 2021-03-21 11:47:51 +0200

tmonteil gravatar image

updated 2021-03-21 16:28:20 +0200

slelievre gravatar image

First, your question adds a useless level of complexity, since everything happens within A[0], so let me first reduce it:

sage: B = [9, 9, 1, 9]
sage: B.index(B[0])
0
sage: B.index(B[1])
0
sage: B.index(B[2])
2
sage: B.index(B[3])
0

You can check the documentation of the index method of lists as follows:

sage: B.index?

As you can see, this method B.index(x) returns the smallest index i such that B[i]=x (when it exists, otherwise it raises a ValueError).

In our case, there are three indices i for which B[i] equals 9, namely 0, 1, 3. The smallest of them is 0, hence B.index(9) will return 0. The fact that 9 was provided by B[0], B[1] or B[3] does not matter.

Note that this question is only about the Python language, not Sage. Hence you might find tons of explanations about this on the web.

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Sorry to both. As I am not a professional in programming some times, I lose control.

Cyrille gravatar imageCyrille ( 2021-03-22 12:15:36 +0200 )edit

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Asked: 2021-03-21 11:27:15 +0200

Seen: 44 times

Last updated: Mar 21