Hi I am trying to plot a numpy array of strings in y axis, for example
arr = np.array(['a','a','bas','dgg','a']) #The actual strings are about 11 characters long
vs a float array with equal length. The string array I am working with is very large ~ 100 million entries. One of the solutions I had in mind was to convert the string array to unique integer ids, for example,
vocab = np.unique(arr)
vocab = list(vocab)
arrId = np.zeros(len(arr))
for i in range(len(arr)):
arrId[i] = vocab.index(arr[i])
and then matplotlib.pyplot.plot(arrId). But I cannot afford to run a for loop to convert the array of strings to an array of unique integer ids. In an initial search I could not find a way to map strings to an unique id without using a loop. Maybe I am missing something, but is there a smart way to do this in python?
EDIT -
Thanks. The solutions provided use vocab,ind = np.unique(arr, return_index = True) where idx is the returned unique integer array. But it seems like np.unique is O(N*log(N)) according to this ( numpy.unique with order preserved), but pandas.unique is of order O(N). But I am not sure how to get ind from pandas.unique. plotting data i guess can be done in O(N). So I was wondering is there a way to do this O(N)? perhaps by hashing of some sort?
numpy.unique used with the return_inverse argument allows you to obtain the inverted index.
arr = np.array(['a','a','bas','dgg','a'])
unique, rev = np.unique(arr, return_inverse=True)
#unique: ['a' 'bas' 'dgg']
#rev: [0 0 1 2 0]
such that unique[rev] returns the original array ['a' 'a' 'bas' 'dgg' 'a'].
This can be easily used to plot the data.
import numpy as np
import matplotlib.pyplot as plt
arr = np.array(['a','a','bas','dgg','a'])
x = np.array([1,2,3,4,5])
unique, rev = np.unique(arr, return_inverse=True)
print unique
print rev
print unique[rev]
fig,ax=plt.subplots()
ax.scatter(x, rev)
ax.set_yticks(range(len(unique)))
ax.set_yticklabels(unique)
plt.show()
you can factorize your strings:
In [75]: arr = np.array(['a','a','bas','dgg','a'])
In [76]: cats, idx = np.unique(arr, return_inverse=True)
In [77]: plt.plot(idx)
Out[77]: [<matplotlib.lines.Line2D at 0xf82da58>]
In [78]: cats
Out[78]:
array(['a', 'bas', 'dgg'],
dtype='<U3')
In [79]: idx
Out[79]: array([0, 0, 1, 2, 0], dtype=int64)
You can use the numpy unique funciton to return a unique array of values?
print(np.unique(arr))
['a' 'bas' 'dgg']
collections.counter also return the value and number of counts:
print(collections.Counter(arr))
Counter({'a': 3, 'bas': 1, 'dgg': 1})
Does this help at all?
This question already has answers here:
Is there a NumPy function to return the first index of something in an array?
(20 answers)
Closed 2 years ago.
In Python we can get the index of a value in an array by using .index().
But with a NumPy array, when I try to do:
decoding.index(i)
I get:
AttributeError: 'numpy.ndarray' object has no attribute 'index'
How could I do this on a NumPy array?
Use np.where to get the indices where a given condition is True.
Examples:
For a 2D np.ndarray called a:
i, j = np.where(a == value) # when comparing arrays of integers
i, j = np.where(np.isclose(a, value)) # when comparing floating-point arrays
For a 1D array:
i, = np.where(a == value) # integers
i, = np.where(np.isclose(a, value)) # floating-point
Note that this also works for conditions like >=, <=, != and so forth...
You can also create a subclass of np.ndarray with an index() method:
class myarray(np.ndarray):
def __new__(cls, *args, **kwargs):
return np.array(*args, **kwargs).view(myarray)
def index(self, value):
return np.where(self == value)
Testing:
a = myarray([1,2,3,4,4,4,5,6,4,4,4])
a.index(4)
#(array([ 3, 4, 5, 8, 9, 10]),)
You can convert a numpy array to list and get its index .
for example:
tmp = [1,2,3,4,5] #python list
a = numpy.array(tmp) #numpy array
i = list(a).index(2) # i will return index of 2, which is 1
this is just what you wanted.
I'm torn between these two ways of implementing an index of a NumPy array:
idx = list(classes).index(var)
idx = np.where(classes == var)
Both take the same number of characters, but the first method returns an int instead of a numpy.ndarray.
This problem can be solved efficiently using the numpy_indexed library (disclaimer: I am its author); which was created to address problems of this type. npi.indices can be viewed as an n-dimensional generalisation of list.index. It will act on nd-arrays (along a specified axis); and also will look up multiple entries in a vectorized manner as opposed to a single item at a time.
a = np.random.rand(50, 60, 70)
i = np.random.randint(0, len(a), 40)
b = a[i]
import numpy_indexed as npi
assert all(i == npi.indices(a, b))
This solution has better time complexity (n log n at worst) than any of the previously posted answers, and is fully vectorized.
You can use the function numpy.nonzero(), or the nonzero() method of an array
import numpy as np
A = np.array([[2,4],
[6,2]])
index= np.nonzero(A>1)
OR
(A>1).nonzero()
Output:
(array([0, 1]), array([1, 0]))
First array in output depicts the row index and second array depicts the corresponding column index.
If you are interested in the indexes, the best choice is np.argsort(a)
a = np.random.randint(0, 100, 10)
sorted_idx = np.argsort(a)
Suppose you have a numpy array and a list:
>>> a = np.array([1,2,2,1]).reshape(2,2)
>>> a
array([[1, 2],
[2, 1]])
>>> b = [0, 10]
I'd like to replace values in an array, so that 1 is replaced by 0, and 2 by 10.
I found a similar problem here - http://mail.python.org/pipermail//tutor/2011-September/085392.html
But using this solution:
for x in np.nditer(a):
if x==1:
x[...]=x=0
elif x==2:
x[...]=x=10
Throws me an error:
ValueError: assignment destination is read-only
I guess that's because I can't really write into a numpy array.
P.S. The actual size of the numpy array is 514 by 504 and of the list is 8.
Well, I suppose what you need is
a[a==2] = 10 #replace all 2's with 10's
Instead of replacing the values one by one, it is possible to remap the entire array like this:
import numpy as np
a = np.array([1,2,2,1]).reshape(2,2)
# palette must be given in sorted order
palette = [1, 2]
# key gives the new values you wish palette to be mapped to.
key = np.array([0, 10])
index = np.digitize(a.ravel(), palette, right=True)
print(key[index].reshape(a.shape))
yields
[[ 0 10]
[10 0]]
Credit for the above idea goes to #JoshAdel. It is significantly faster than my original answer:
import numpy as np
import random
palette = np.arange(8)
key = palette**2
a = np.array([random.choice(palette) for i in range(514*504)]).reshape(514,504)
def using_unique():
palette, index = np.unique(a, return_inverse=True)
return key[index].reshape(a.shape)
def using_digitize():
index = np.digitize(a.ravel(), palette, right=True)
return key[index].reshape(a.shape)
if __name__ == '__main__':
assert np.allclose(using_unique(), using_digitize())
I benchmarked the two versions this way:
In [107]: %timeit using_unique()
10 loops, best of 3: 35.6 ms per loop
In [112]: %timeit using_digitize()
100 loops, best of 3: 5.14 ms per loop
Read-only array in numpy can be made writable:
nArray.flags.writeable = True
This will then allow assignment operations like this one:
nArray[nArray == 10] = 9999 # replace all 10's with 9999's
The real problem was not assignment itself but the writable flag.
I found another solution with the numpy function place. (Documentation here)
Using it on your example:
>>> a = np.array([1,2,2,1]).reshape(2,2)
>>> a
array([[1, 2],
[2, 1]])
>>> np.place(a, a==1, 0)
>>> np.place(a, a==2, 10)
>>> a
array([[ 0, 10],
[10, 0]])
You can also use np.choose(idx, vals), where idx is an array of indices that indicate which value of vals should be put in their place. The indices must be 0-based, though. Also make sure that idx has an integer datatype. So you would only need to do:
np.choose(a.astype(np.int32) - 1, b)
I was unable to set the flags, or use a mask to modify the value. In the end I just made a copy of the array.
a2 = np.copy(a)
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()