## For each label in one array set the first k occurrences to False in another array

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I have two (sorted) arrays, A and B, of different lengths each containing unique labels that are repeated a number of times. The count for each label in A is less than or equal to that in B. All labels in A will be in B, but some labels in B do not appear in A.

I need an object the same length as B where, for each label `i` in A (which occurs `k_i` times), the first `k_i` occurrences of label `i` in B need to be set to `False`. The remaining elements should be `True`.

The following code gives me what I need, but if A and B are large, this can take a long time:

```import numpy as np

# The labels and their frequency
A = np.array((1,1,2,2,3,4,4,4))
B = np.array((1,1,1,1,1,2,2,3,3,4,4,4,4,4,5,5))

A_uniq, A_count = np.unique(A, return_counts = True)
new_ind = np.ones(B.shape, dtype = bool)
for i in range(len(A_uniq)):
new_ind[np.where(B == A_uniq[i])[:A_count[i]]] = False

print(new_ind)
#[False False  True  True  True False False False  True False False False
#  True  True  True  True]
```

Is there a faster or more efficient way to do this? I feel like I may be missing some obvious broadcasting or vectorized solution.

Here's one with `np.searchsorted` -

```idx = np.searchsorted(B, A_uniq)
id_ar = np.zeros(len(B),dtype=int)
id_ar[idx] = 1
id_ar[A_count+idx] -= 1
out = id_ar.cumsum()==0
```

We can optimize further to compute `A_uniq,A_count` using its sorted nature instead of using `np.unique`, like so -

```mask_A = np.r_[True,A[:-1]!=A[1:],True]
```

Sort an array according to the order defined by another array , Source: Amazon Interview | Set 110 (On-Campus) Create another array visited​[] and initialize all entries in it as false. visited[] is used to A Binary Search based function to find index of FIRST occurrence be made equal by maximum K increments · Count of subarrays of size K with elements having even frequencies. Here is a way to count occurrences inside an array of objects. It also places the first array's contents inside a new array to sort the values so that the order in the original array is not disrupted. Then a recursive function is used to go through each element and count the quantity property of each object inside the array.

Example without numpy

```A = [1,1,2,2,3,4,4,4]
B = [1,1,1,1,1,2,2,3,3,4,4,4,4,4,5,5]

a_i = b_i = 0
while a_i < len(A):
if A[a_i] == B[b_i]:
a_i += 1
B[b_i] = False
else:
B[b_i] = True
b_i += 1
# fill the rest of B with True
B[b_i:] = [True] * (len(B) - b_i)
# [False, False, True, True, True, False, False, False, True, False, False, False, True, True, True, True]
```

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This solution is inspired by the one by @Divakar, using itertools.groupby:

```import numpy as np
from itertools import groupby
A = np.array((1, 1, 2, 2, 3, 4, 4, 4))
B = np.array((1, 1, 1, 1, 1, 2, 2, 3, 3, 4, 4, 4, 4, 4, 5, 5))

indices = [key + i for key, group in groupby(np.searchsorted(B, A)) for i, _ in enumerate(group)]
result = np.ones_like(B, dtype=np.bool)
result[indices] = False

print(result)
```

Output

```[False False  True  True  True False False False  True False False False
True  True  True  True]
```

The idea is to use np.searchsorted to find the insertion position of each element of `A`, as equal elements will have the same insertion position you have to shift by one each of them, hence the groupby. Then create an array of `True` and set the values of the `indices` to `False`.

If you can use `pandas`, compute the `indices` like this:

```values = np.searchsorted(B, A)
indices = pd.Series(values).groupby(values).cumcount() + values
```

Applied Data Mining, Zhenglu Yang. Let x be an instance, y be the binary label vector associate with x, and N(x) represents its k nearest neighbours in the training set. For each labell, ML-KNN will calculate the following statistics information first of all. For labell(​1sism), there will be an array:k, which has k+1 elements. The value of k's jth  Count the occurrences of an element in an array in Java. We will be performing the below steps to count the occurrence. As a first step we will be creating a HashMap “countMap” to hold the element (Key) and the count as the value. For each of the element in the input array, check if it is present in the countMap, using containsKey() method.

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Combinatorics on Words: Christoffel Words and Repetitions in Words, Given a word w of length n, find all locations of all occurrences of a factor f within w (more precisely, O(nlog|A|) time, but one typically assumes a fixed alphabet). T(w) one can solve the factor problem in O(|f| + k) time, assuming k instances Though admittedly they introduce so-called branching squares and DFS arrays. The fill() method changes all elements in an array to a static value, from a start index (default 0) to an end index (default array.length). It returns the modified array.

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