Python列表。 查找所有元素不为零的最大行

Question

I have a python list which looks a little like this:

[[0,0,0,0,0,0,0,0,0,0,0,0,0,0],[0,0,90,1,9999,0,0,0,0,0,0,0,00,0],
[0,0,0,0,0,0,0,0,0,0,0,0,00,0],[0,0,90,1,9999,1,2,0,0,9999,0,0,00,0].....till about 30 rows]

I need to find the maximum row from this list which has a 9999 or in other words, which does not have all elements zero. Please help me with this. Thanks!!

I tried:

print max((numpy.where(v1==9999)[0])) 

but this just gave me some weird errors like 'int' object not iterable or numpy.where does not accept keywords n and so on!!

Answer 1

Do you want:

idx,row = max(enumerate(lst),key=lambda r: ( sum(r[1])==0, r[0] ) )

where lst is your list.

Or do you want:

next(x for x in reversed(lst) if sum(x) != 0)

Answer 2

    python 3.2

    #   if you want to find how many rows in your list has 9999
    v=[[0,0,0,0,0,0,0,0,0,0,0,0,0,0],[0,0,90,1,9999,0,0,0,0,0,0,0,00,0]]

    len([i for i in v if 9999 in i])




 #   if any element in your rows is 9999 then all the elements of that row 
 # cannot be 0. then if you want to know how many rows have all the elements 0.


     len([i for i in v if sum(i)==0])

Answer 3

This will return the indices that have non-"all-zero" rows:

nonzerorows = [i for i,j in enumerate(a) if any(j)]

And the highest index of such rows:

maxnonzerorows = max([i for i,j in enumerate(a) if any(j)])