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 ?
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.
Asked: 2021-03-21 11:27:15 +0100
Seen: 211 times
Last updated: Mar 21 '21
Copyright Sage, 2010. Some rights reserved under creative commons license. Content on this site is licensed under a Creative Commons Attribution Share Alike 3.0 license.
