A doubt about question No.74

[madeirak]

74. Given an array C that is a bincount, how to produce an array A such that np.bincount(A) == C? (★★★)

The given answer is shown as below.
C = np.bincount([1,1,2,3,4,4,6]) A = np.repeat(np.arange(len(C)), C) print(A)

There maybe a problem with the description of question No.74,the answer given can only solve the problem of Non-strictly increasing 1-d array not for all kind of 1-d array.

[rougier]

Do you have an example where this fails?

[madeirak]

Do you have an example where this fails?

1. problem
   The example in given answer is created by direct assignment,not by common means like np.random.randint(0,10,7)
   The correct example must be a Non-strictly increasing 1-d array like the answer given [1,1,2,3,4,4,6],whose the next adjacent element is larger than the previous one.

A counterexample is like [2,1,1,1,1,1,1]

C = np.bincount([2,1,1,1,1,1,1,1])
A = np.repeat(np.arange(len(C)), C)
#print C and A
#C= [0 7 1]
#A= [1 1 1 1 1 1 1 2]

We can see [2,1,1,1,1,1,1,1] is not equal to [1,1,1,1,1,1,1,2],we get the wrong answer.
The key point is np.repeat() is sequential.

2. Possible solutions
   Adding the notion of non-strictly increasing 1-d array in the title description of No.74 is the most convenient way.

[rougier]

Thanks for the ifnromation. We could simply add sorted in C description: "Given a sorted array C that corresponds to a bincount". Can you make a PR?

[madeirak]

Thanks for the ifnromation. We could simply add sorted in C description: "Given a sorted array C that corresponds to a bincount". Can you make a PR?

It`s my pleasure :)

[madeirak]

#138

[rougier]

Thanks, merged.