I am very new to python. I want to clearly understand the below code, if there's anyone who can help me.
Code:
import numpy as np
arr = np.array([[1, 2, 3, 4,99,11,22], [5, 6, 7, 8,43,54,22]])
for x in np.nditer(arr[0:,::4]):
print(x)
My understanding:
This 2D array has two 1D arrays.
np.nditer(arr[0:,::4]) will give all value from 0 indexed array to upto last array, ::4 means the gap between printed arrays will be 4.
Question:
Is my understanding for no 2 above correct?
How can I get the index for the print(x)? Because of the step difference of 4 e.g [0:,::4] or any gap [0:,::x] I want to find out the exact index that it is printing. But how?
Addressing your questions below
Yes, I think your understanding is correct. It might help to first print what arr[0:,::4] returns though:
iter_array = arr[0:,::4]
print(iter_array)
>>> [[ 1 99]
>>> [ 5 43]]
The slicing takes out each 4th index of the original array. All nditer does is iterate through these values in order. (Quick FYI: arr[0:] and arr[:] are equivalent, since the starting point is 0 by default).
As you pointed out, to get the index for these you need to keep track of the slicing that you did, i.e. arr[0:, ::x]. Remember, nditer has nothing to do with how you sliced your array. I'm not sure how to best get the indices of your slicing, but this is what I came up with:
import numpy as np
ls = [
[1, 2, 3, 4,99,11,22],
[5, 6, 7, 8,43,54,22]
]
arr = np.array(ls)
inds = np.array([
[(ctr1, ctr2) for ctr2, _ in enumerate(l)] for ctr1, l in enumerate(ls)
]) # create duplicate of arr filled with zeros
step = 4
iter_array = arr[0:,::step]
iter_inds = inds[0:,::step]
print(iter_array)
>>> [[ 1 99]
>>> [ 5 43]]
print(iter_inds)
>>> [[[0 0]
>>> [0 4]]
>>>
>>> [[1 0]
>>> [1 4]]]
All that I added here was an inds array. This array has elements equal to their own index. Then, when you slice both arrays in the same way, you get your indices. Hopefully this helps!
One interesting question:
I would like to delete some elements from a numpy array but just as below simplified example code, it works if didn't delete the last element, but it failure if we wish to delete the last element.
Below code works fine:
import numpy as np
values = np.array([0,1,2,3,4,5])
print values
for i in [3,4,1]:
values = np.delete(values,i)
print values
The output is:
[0 1 2 3 4 5]
[0 2 4]
If we only change 4 to 5, then it will fail:
import numpy as np
values = np.array([0,1,2,3,4,5])
print values
for i in [3,5,1]:
values = np.delete(values,i)
print values
The error message:
IndexError: index 5 is out of bounds for axis 0 with size 5
Why this error only happen if delete the last element? what's correct way to do such tasks?
Keep in mind that np.delete(arr, ind) deletes the element at index ind NOT the one with that value.
This means that as you delete things, the array is getting shorter. So you start with
values = [0,1,2,3,4,5]
np.delete(values, 3)
[0,1,2,4,5] #deleted element 3 so now only 5 elements in the list
#tries to delete the element at the fifth index but the array indices only go from 0-4
np.delete(values, 5)
One of the ways you can solve the problem is to sort the indices that you want to delete in descending order (if you really want to delete the array).
inds_to_delete = sorted([3,1,5], reverse=True) # [5,3,1]
# then delete in order of largest to smallest ind
Or:
inds_to_keep = np.array([0,2,4])
values = values[inds_to_keep]
A probably faster way (because you don't need to delete every single value but all at once) is using a boolean mask:
values = np.array([0,1,2,3,4,5])
tobedeleted = np.array([False, True, False, True, False, True])
# So index 3, 5 and 1 are True so they will be deleted.
values_deleted = values[~tobedeleted]
#that just gives you what you want.
It is recommended on the numpy reference on np.delete
To your question: You delete one element so the array get's shorter and index 5 is no longer in the array because the former index 5 has now index 4. Delete in descending order if you want to use np.delete.
If you really want to delete with np.delete use the shorthand:
np.delete(values, [3,5,1])
If you want to delete where the values are (not the index) you have to alter the procedure a bit. If you want to delete all values 5 in your array you can use:
values[values != 5]
or with multiple values to delete:
to_delete = (values == 5) | (values == 3) | (values == 1)
values[~to_delete]
all of these give you the desired result, not sure how your data really looks like so I can't say for sure which will be the most appropriate.
The problem is that you have deleted items from values so when you are trying to delete item in index 5 there is no longer value at that index, it's now at index 4.
If you sort the list of indices to delete, and iterate over them from large to small that should workaround this issue.
import numpy as np
values = np.array([0,1,2,3,4,5])
print values
for i in [5,3,1]: # iterate in order
values = np.delete(values,i)
print values
If you want to remove the elements of indices 3,4,1 , just do np.delete(values,[3,4,1]).
If you want in the first case to delete the fourth (index=3) item, then the fifth of the rest and finally the second of the rest, due to the order of the operations, you delete the second, fourth and sixth of the initial array. It's therefore logic that the second case fails.
You can compute the shifts (in the exemple fifth become sixth) in this way :
def multidelete(values,todelete):
todelete=np.array(todelete)
shift=np.triu((todelete>=todelete[:,None]),1).sum(0)
return np.delete(values,todelete+shift)
Some tests:
In [91]: multidelete([0, 1, 2, 3, 4, 5],[3,4,1])
Out[91]: array([0, 2, 4])
In [92]: multidelete([0, 1, 2, 3, 4, 5],[1,1,1])
Out[92]: array([0, 4, 5])
N.B. np.delete doesn't complain an do nothing if the bad indice(s) are in a list : np.delete(values,[8]) is values .
Boolean index is deprected. You can use function np.where() instead like this:
values = np.array([0,1,2,3,4,5])
print(values)
for i in [3,5,1]:
values = np.delete(values,np.where(values==i))
# values = np.delete(values,values==i) # still works with warning
print(values)
I know this question is old, but for further reference (as I found a similar source problem):
Instead of making a for loop, a solution is to filter the array with isin numpy's function. Like so,
>>> import numpy as np
>>> # np.isin(element, test_elements, assume_unique=False, invert=False)
>>> arr = np.array([1, 4, 7, 10, 5, 10])
>>> ~np.isin(arr, [4, 10])
array([ True, False, True, False, True, False])
>>> arr = arr[ ~np.isin(arr, [4, 10]) ]
>>> arr
array([1, 7, 5])
So for this particular case we can write:
values = np.array([0,1,2,3,4,5])
torem = [3,4,1]
values = values[ ~np.isin(values, torem) ]
which outputs: array([0, 2, 5])
here's how you can do it without any loop or any indexing, using numpy.setdiff1d
>>> import numpy as np
>>> array_1 = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])
>>> array_1
array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10])
>>> remove_these = np.array([1,3,5,7,9])
>>> remove_these
array([1, 3, 5, 7, 9])
>>> np.setdiff1d(array_1, remove_these)
array([ 2, 4, 6, 8, 10])
I know there is a method for a Python list to return the first index of something:
>>> xs = [1, 2, 3]
>>> xs.index(2)
1
Is there something like that for NumPy arrays?
Yes, given an array, array, and a value, item to search for, you can use np.where as:
itemindex = numpy.where(array == item)
The result is a tuple with first all the row indices, then all the column indices.
For example, if an array is two dimensions and it contained your item at two locations then
array[itemindex[0][0]][itemindex[1][0]]
would be equal to your item and so would be:
array[itemindex[0][1]][itemindex[1][1]]
If you need the index of the first occurrence of only one value, you can use nonzero (or where, which amounts to the same thing in this case):
>>> t = array([1, 1, 1, 2, 2, 3, 8, 3, 8, 8])
>>> nonzero(t == 8)
(array([6, 8, 9]),)
>>> nonzero(t == 8)[0][0]
6
If you need the first index of each of many values, you could obviously do the same as above repeatedly, but there is a trick that may be faster. The following finds the indices of the first element of each subsequence:
>>> nonzero(r_[1, diff(t)[:-1]])
(array([0, 3, 5, 6, 7, 8]),)
Notice that it finds the beginning of both subsequence of 3s and both subsequences of 8s:
[1, 1, 1, 2, 2, 3, 8, 3, 8, 8]
So it's slightly different than finding the first occurrence of each value. In your program, you may be able to work with a sorted version of t to get what you want:
>>> st = sorted(t)
>>> nonzero(r_[1, diff(st)[:-1]])
(array([0, 3, 5, 7]),)
You can also convert a NumPy array to list in the air and get its index. For example,
l = [1,2,3,4,5] # Python list
a = numpy.array(l) # NumPy array
i = a.tolist().index(2) # i will return index of 2
print i
It will print 1.
Just to add a very performant and handy numba alternative based on np.ndenumerate to find the first index:
from numba import njit
import numpy as np
#njit
def index(array, item):
for idx, val in np.ndenumerate(array):
if val == item:
return idx
# If no item was found return None, other return types might be a problem due to
# numbas type inference.
This is pretty fast and deals naturally with multidimensional arrays:
>>> arr1 = np.ones((100, 100, 100))
>>> arr1[2, 2, 2] = 2
>>> index(arr1, 2)
(2, 2, 2)
>>> arr2 = np.ones(20)
>>> arr2[5] = 2
>>> index(arr2, 2)
(5,)
This can be much faster (because it's short-circuiting the operation) than any approach using np.where or np.nonzero.
However np.argwhere could also deal gracefully with multidimensional arrays (you would need to manually cast it to a tuple and it's not short-circuited) but it would fail if no match is found:
>>> tuple(np.argwhere(arr1 == 2)[0])
(2, 2, 2)
>>> tuple(np.argwhere(arr2 == 2)[0])
(5,)
l.index(x) returns the smallest i such that i is the index of the first occurrence of x in the list.
One can safely assume that the index() function in Python is implemented so that it stops after finding the first match, and this results in an optimal average performance.
For finding an element stopping after the first match in a NumPy array use an iterator (ndenumerate).
In [67]: l=range(100)
In [68]: l.index(2)
Out[68]: 2
NumPy array:
In [69]: a = np.arange(100)
In [70]: next((idx for idx, val in np.ndenumerate(a) if val==2))
Out[70]: (2L,)
Note that both methods index() and next return an error if the element is not found. With next, one can use a second argument to return a special value in case the element is not found, e.g.
In [77]: next((idx for idx, val in np.ndenumerate(a) if val==400),None)
There are other functions in NumPy (argmax, where, and nonzero) that can be used to find an element in an array, but they all have the drawback of going through the whole array looking for all occurrences, thus not being optimized for finding the first element. Note also that where and nonzero return arrays, so you need to select the first element to get the index.
In [71]: np.argmax(a==2)
Out[71]: 2
In [72]: np.where(a==2)
Out[72]: (array([2], dtype=int64),)
In [73]: np.nonzero(a==2)
Out[73]: (array([2], dtype=int64),)
Time comparison
Just checking that for large arrays the solution using an iterator is faster when the searched item is at the beginning of the array (using %timeit in the IPython shell):
In [285]: a = np.arange(100000)
In [286]: %timeit next((idx for idx, val in np.ndenumerate(a) if val==0))
100000 loops, best of 3: 17.6 µs per loop
In [287]: %timeit np.argmax(a==0)
1000 loops, best of 3: 254 µs per loop
In [288]: %timeit np.where(a==0)[0][0]
1000 loops, best of 3: 314 µs per loop
This is an open NumPy GitHub issue.
See also: Numpy: find first index of value fast
If you're going to use this as an index into something else, you can use boolean indices if the arrays are broadcastable; you don't need explicit indices. The absolute simplest way to do this is to simply index based on a truth value.
other_array[first_array == item]
Any boolean operation works:
a = numpy.arange(100)
other_array[first_array > 50]
The nonzero method takes booleans, too:
index = numpy.nonzero(first_array == item)[0][0]
The two zeros are for the tuple of indices (assuming first_array is 1D) and then the first item in the array of indices.
For one-dimensional sorted arrays, it would be much more simpler and efficient O(log(n)) to use numpy.searchsorted which returns a NumPy integer (position). For example,
arr = np.array([1, 1, 1, 2, 3, 3, 4])
i = np.searchsorted(arr, 3)
Just make sure the array is already sorted
Also check if returned index i actually contains the searched element, since searchsorted's main objective is to find indices where elements should be inserted to maintain order.
if arr[i] == 3:
print("present")
else:
print("not present")
For 1D arrays, I'd recommend np.flatnonzero(array == value)[0], which is equivalent to both np.nonzero(array == value)[0][0] and np.where(array == value)[0][0] but avoids the ugliness of unboxing a 1-element tuple.
To index on any criteria, you can so something like the following:
In [1]: from numpy import *
In [2]: x = arange(125).reshape((5,5,5))
In [3]: y = indices(x.shape)
In [4]: locs = y[:,x >= 120] # put whatever you want in place of x >= 120
In [5]: pts = hsplit(locs, len(locs[0]))
In [6]: for pt in pts:
.....: print(', '.join(str(p[0]) for p in pt))
4, 4, 0
4, 4, 1
4, 4, 2
4, 4, 3
4, 4, 4
And here's a quick function to do what list.index() does, except doesn't raise an exception if it's not found. Beware -- this is probably very slow on large arrays. You can probably monkey patch this on to arrays if you'd rather use it as a method.
def ndindex(ndarray, item):
if len(ndarray.shape) == 1:
try:
return [ndarray.tolist().index(item)]
except:
pass
else:
for i, subarray in enumerate(ndarray):
try:
return [i] + ndindex(subarray, item)
except:
pass
In [1]: ndindex(x, 103)
Out[1]: [4, 0, 3]
An alternative to selecting the first element from np.where() is to use a generator expression together with enumerate, such as:
>>> import numpy as np
>>> x = np.arange(100) # x = array([0, 1, 2, 3, ... 99])
>>> next(i for i, x_i in enumerate(x) if x_i == 2)
2
For a two dimensional array one would do:
>>> x = np.arange(100).reshape(10,10) # x = array([[0, 1, 2,... 9], [10,..19],])
>>> next((i,j) for i, x_i in enumerate(x)
... for j, x_ij in enumerate(x_i) if x_ij == 2)
(0, 2)
The advantage of this approach is that it stops checking the elements of the array after the first match is found, whereas np.where checks all elements for a match. A generator expression would be faster if there's match early in the array.
There are lots of operations in NumPy that could perhaps be put together to accomplish this. This will return indices of elements equal to item:
numpy.nonzero(array - item)
You could then take the first elements of the lists to get a single element.
Comparison of 8 methods
TL;DR:
(Note: applicable to 1d arrays under 100M elements.)
For maximum performance use index_of__v5 (numba + numpy.enumerate + for loop; see the code below).
If numba is not available:
Use index_of__v7 (for loop + enumerate) if the target value is expected to be found within the first 100k elements.
Else use index_of__v2/v3/v4 (numpy.argmax or numpy.flatnonzero based).
Powered by perfplot
import numpy as np
from numba import njit
# Based on: numpy.argmax()
# Proposed by: John Haberstroh (https://stackoverflow.com/a/67497472/7204581)
def index_of__v1(arr: np.array, v):
is_v = (arr == v)
return is_v.argmax() if is_v.any() else -1
# Based on: numpy.argmax()
def index_of__v2(arr: np.array, v):
return (arr == v).argmax() if v in arr else -1
# Based on: numpy.flatnonzero()
# Proposed by: 1'' (https://stackoverflow.com/a/42049655/7204581)
def index_of__v3(arr: np.array, v):
idxs = np.flatnonzero(arr == v)
return idxs[0] if len(idxs) > 0 else -1
# Based on: numpy.argmax()
def index_of__v4(arr: np.array, v):
return np.r_[False, (arr == v)].argmax() - 1
# Based on: numba, for loop
# Proposed by: MSeifert (https://stackoverflow.com/a/41578614/7204581)
#njit
def index_of__v5(arr: np.array, v):
for idx, val in np.ndenumerate(arr):
if val == v:
return idx[0]
return -1
# Based on: numpy.ndenumerate(), for loop
def index_of__v6(arr: np.array, v):
return next((idx[0] for idx, val in np.ndenumerate(arr) if val == v), -1)
# Based on: enumerate(), for loop
# Proposed by: Noyer282 (https://stackoverflow.com/a/40426159/7204581)
def index_of__v7(arr: np.array, v):
return next((idx for idx, val in enumerate(arr) if val == v), -1)
# Based on: list.index()
# Proposed by: Hima (https://stackoverflow.com/a/23994923/7204581)
def index_of__v8(arr: np.array, v):
l = list(arr)
try:
return l.index(v)
except ValueError:
return -1
Go to Colab
The numpy_indexed package (disclaimer, I am its author) contains a vectorized equivalent of list.index for numpy.ndarray; that is:
sequence_of_arrays = [[0, 1], [1, 2], [-5, 0]]
arrays_to_query = [[-5, 0], [1, 0]]
import numpy_indexed as npi
idx = npi.indices(sequence_of_arrays, arrays_to_query, missing=-1)
print(idx) # [2, -1]
This solution has vectorized performance, generalizes to ndarrays, and has various ways of dealing with missing values.
There is a fairly idiomatic and vectorized way to do this built into numpy. It uses a quirk of the np.argmax() function to accomplish this -- if many values match, it returns the index of the first match. The trick is that for booleans, there will only ever be two values: True (1) and False (0). Therefore, the returned index will be that of the first True.
For the simple example provided, you can see it work with the following
>>> np.argmax(np.array([1,2,3]) == 2)
1
A great example is computing buckets, e.g. for categorizing. Let's say you have an array of cut points, and you want the "bucket" that corresponds to each element of your array. The algorithm is to compute the first index of cuts where x < cuts (after padding cuts with np.Infitnity). I could use broadcasting to broadcast the comparisons, then apply argmax along the cuts-broadcasted axis.
>>> cuts = np.array([10, 50, 100])
>>> cuts_pad = np.array([*cuts, np.Infinity])
>>> x = np.array([7, 11, 80, 443])
>>> bins = np.argmax( x[:, np.newaxis] < cuts_pad[np.newaxis, :], axis = 1)
>>> print(bins)
[0, 1, 2, 3]
As expected, each value from x falls into one of the sequential bins, with well-defined and easy to specify edge case behavior.
Another option not previously mentioned is the bisect module, which also works on lists, but requires a pre-sorted list/array:
import bisect
import numpy as np
z = np.array([104,113,120,122,126,138])
bisect.bisect_left(z, 122)
yields
3
bisect also returns a result when the number you're looking for doesn't exist in the array, so that the number can be inserted in the correct place.
Note: this is for python 2.7 version
You can use a lambda function to deal with the problem, and it works both on NumPy array and list.
your_list = [11, 22, 23, 44, 55]
result = filter(lambda x:your_list[x]>30, range(len(your_list)))
#result: [3, 4]
import numpy as np
your_numpy_array = np.array([11, 22, 23, 44, 55])
result = filter(lambda x:your_numpy_array [x]>30, range(len(your_list)))
#result: [3, 4]
And you can use
result[0]
to get the first index of the filtered elements.
For python 3.6, use
list(result)
instead of
result
Use ndindex
Sample array
arr = np.array([[1,4],
[2,3]])
print(arr)
...[[1,4],
[2,3]]
create an empty list to store the index and the element tuples
index_elements = []
for i in np.ndindex(arr.shape):
index_elements.append((arr[i],i))
convert the list of tuples into dictionary
index_elements = dict(index_elements)
The keys are the elements and the values are their
indices - use keys to access the index
index_elements[4]
output
... (0,1)
For my use case, I could not sort the array ahead of time because the order of the elements is important. This is my all-NumPy implementation:
import numpy as np
# The array in question
arr = np.array([1,2,1,2,1,5,5,3,5,9])
# Find all of the present values
vals=np.unique(arr)
# Make all indices up-to and including the desired index positive
cum_sum=np.cumsum(arr==vals.reshape(-1,1),axis=1)
# Add zeros to account for the n-1 shape of diff and the all-positive array of the first index
bl_mask=np.concatenate([np.zeros((cum_sum.shape[0],1)),cum_sum],axis=1)>=1
# The desired indices
idx=np.where(np.diff(bl_mask))[1]
# Show results
print(list(zip(vals,idx)))
>>> [(1, 0), (2, 1), (3, 7), (5, 5), (9, 9)]
I believe it accounts for unsorted arrays with duplicate values.
Found another solution with loops:
new_array_of_indicies = []
for i in range(len(some_array)):
if some_array[i] == some_value:
new_array_of_indicies.append(i)
index_lst_form_numpy = pd.DataFrame(df).reset_index()["index"].tolist()