Say I have an array of values nameArr = ['josh','is','a','person'] and I want a function like arrayLocation(nameArr,['a','is']) to return [2 1].
Does that function already exist, if not how can I implement it efficiently?
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4 Answers
use numpy.where
In [17]: nameArr = np.array(['josh','is','a','person'])
In [18]: [np.where(nameArr==i) for i in ['a','is']]
Out[18]: [(array([2]),), (array([1]),)]
Comments
Lists have a index method that you can use.
>>> nameArr = ['josh','is','a','person']
>>> # Using map
>>> map(nameArr.index, ['a', 'is'])
[2, 1]
>>> # Using list comprehensions
>>> [nameArr.index(x) for x in ['a', 'is']]
[2, 1]
BTW, index raises ValueError if the element is not in the list. So if you want to supply elements that are not in the list to the index method, you may need to handle the error appropriately.
answered Apr 20, 2012 at 20:02
Praveen Gollakota
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1 Comment
agf
Also, it always returns the first index of the item, if the item is in the list multiple times.
If the array is large, and do many times of location, you can create a dict that map the value to it's index first:
d = dict(zip(nameArr, range(len(nameArr))))
items = ['a','is']
print [d.get(x, None) for x in items]
1 Comment
Grr
Given scalability as a measure of quality this is by far the best answer here.
Something like:
>>> f_idxs = np.ravel([np.where(master_data==i)[0] for i in search_list])
