I know how can I find the number of duplicates from a numpy array. However, I need to find the number of duplicates only at the end of the numpy array. Please see my example below:
Example input is as follows:
1995
1996
1996
1997
1998
1999
1999
1999
Desired output:
3
Thanks in advance!
Here's one way with np.minimum.accumulate -
np.minimum.accumulate(a[::-1]==a[-1]).sum()
Sample run -
In [64]: a
Out[64]: array([2, 1, 9, 0, 0, 0, 2, 1, 0, 0, 0, 0, 2, 1, 0, 9, 9, 9, 9, 9])
In [73]: np.minimum.accumulate(a[::-1]==a[-1]).sum()
Out[73]: 5
Another with argmin -
In [88]: (a[::-1]==a[-1]).argmin()
Out[88]: 5
For a corner case, if all elements are same, we might need one extra step to check for all matches on a[::-1]==a[-1] and return len(a) in that case. Or if the count is 0, which can't be the output, we will output len(a) instead.
only at the end of the numpy array correctly.
mylist = [1995,
1996,
1996,
1997,
1998,
1999,
1999,
1999]
s = 0
x = mylist[-1]
for i in mylist:
if i == x: s+= 1
print(s)
My solution is as follows, however, I am sure that there is a more elegant solution in a line!
def get_numlastdups(my_nparray):
numlastdups=0
for i in range(2,len(my_nparray)):
if my_nparray[-1]==my_nparray[-i]: numlastdups+=1
else: break
return numlastdups
