#!/usr/bin/env python
# coding: utf-8

# ## HLL Sketch Examples

# ### Basic Sketch Usage

# In[1]:


from datasketches import hll_sketch, hll_union, tgt_hll_type


# We'll create a sketch with log2(k) = 12

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sk = hll_sketch(12)


# Insert ~2 million points. Values are hashed, so using sequential integers is fine for demonstration purposes.

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n = 1 << 21
for i in range(0, n):
    sk.update(i)
print(sk)


# Since we know the exact value of n we can look at the estimate and upper/lower bounds as a % of the true value. We'll look at the bounds at 1 standard deviation. In this case, the true value does lie within the bounds, but since these are probabilistic bounds the true value will sometimes be outside them (especially at 1 standard deviation).

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print("Upper bound (1 std. dev) as % of true value: ", round(100*sk.get_upper_bound(1) / n, 4))


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print("Estimate as % of true value: ", round(100*sk.get_estimate() / n, 4))


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print("Lower bound (1 std. dev) as % of true value: ", round(100*sk.get_lower_bound(1) / n, 4))


# Finally, we can serialize and deserialize the sketch, which will give us back the same structure.

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sk_bytes = sk.serialize_compact()
len(sk_bytes)


# In[8]:


sk2 = hll_sketch.deserialize(sk_bytes)
print(sk2)


# ### Sketch Union Usage

# Here, we'll create two sketches with partial overlap in values. For good measure, we'll let k be larger in one sketch. For most applications we'd generally create all new data using the same size sketch, allowing differences to creep in when combining new and historica data.

# In[9]:


k = 12
n = 1 << 20
offset = int(3 * n / 4)


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sk1 = hll_sketch(k)
sk2 = hll_sketch(k + 1)
for i in range(0, n):
    sk1.update(i)
    sk2.update(i + offset)


# Create a union object and add the sketches to that. To demonstrate smoothly handling multiple sketch sizes, we'll use a size of k+1 here.

# In[11]:


union = hll_union(k+1)
union.update(sk1)
union.update(sk2)


# Note how log config k has automatically adopted the value of the smaller input sketch.

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result = union.get_result()
print(result)


# We can again compare against the exact result, in this case 1.75*n

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print("Estimate as % of true value: ", round(100*result.get_estimate() / (7*n/4), 4))


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