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Given a group of values, the entropy of the group is defined as the formula as following:

where `P(x)`

is the probability of appearance for the value `x`

.

*e.g.*

the input group: `[1, 1, 2, 2]`

the probability of value 1 is: 2/4 = 1/2

the probability of value 2 is: 2/4 = 1/2

Therefore, its entropy can be obtained by: `- (1/2) * log`

_{2}(1/2) - (1/2) * log_{2}(1/2) = 1/2 + 1/2 = 1

This exercise, however, is aimed to calculate the maximum `information gain`

that one can obtain by splitting a group into two subgroups. The information gain is the difference of entropy before and after the splitting.

For a group of *L*, we divide it into subgroups of *{L _{1}, L_{2}}*, then the information gain is calculated as following:

The overall entropy of the splitted subgroups *{L _{1}, L_{2}}* is the sum of entropy for each subgroup weighted by its proportion with regards to the original group.

In this exercise, one can expect a list of samples on Iris flowers. Each sample is represented with a tuple of two values: `<petal_length, species>`

, where the first attribute is the measurement on the length of the petal for the sample, and the second attribute indicates the species of sample. Here is an example.

The task is to split the sample list into two sublists, while complying with the following two conditions:

- The
`petal_length`

of sample in one sublist is less or equal than that of any sample in the other sublist. - The information gain on the
`species`

attribute of the sublists is**maximal**among all possible splits.

As output, one should just return the information gain.

In addition, one can expect that each value of `petal_length`

is unique.

In the above example, the optimal split would be *L _{1}* = [(0.5, ‘setosa’), (1.0, ‘setosa’)] and

** Note:** For certain languages (

`[petal_length]`

`[species]`

respectively, where the petal_length would be of float value and the species is of string. The elements in the petal_length list and species list are associated to each other one by one by order.```
class Solution:
def calculateMaxInfoGain(self, petal_length: List[float], species: List[str]) -> float:
if not petal_length or not species:
return 0
n = len(petal_length)
Hs = math.inf
mp = sorted(zip(petal_length, species))
petal_length, species = zip(*mp)
for i in range(1, n):
H1 = self.calculateEntropy(species[:i])
H2 = self.calculateEntropy(species[i:])
Hs = min(Hs, H1*i/n + H2*(n-i)/n)
Hl = self.calculateEntropy(species)
return Hl-Hs
def calculateEntropy(self, species: List[str]) -> float:
size = len(species)
probs = collections.defaultdict(int)
for sp in species:
probs[sp] += 1/size
if len(probs.values())<=1:
return 0
return -sum([p*log2(p) for p in probs.values()])
```

```
class Solution {
public:
double calEntropy(vector<string>& species)
{
unordered_map<string, int> shown;
double total = species.size();
for(const string& specie:species)
{
++shown[specie];
}
double entropy = 0.0;
for(auto& n:shown)
{
double p = n.second / total;
entropy -= p*log(p)/log(2.0);
}
return entropy;
}
double calculateMaxInfoGain(vector<double>& petal_length, vector<string>& species) {
// Avoid edge case
if(petal_length.empty() || species.empty()) return 0.0;
int s = petal_length.size();
double all_entropy = calEntropy(species);
double information_gain = 0.0;
// Iterate all possible splits
for(int i = 0; i < s; ++i)
{
// Split list
double split = petal_length[i];
vector<string> l1;
vector<string> l2;
for(int i = 0; i < s; ++i)
{
if(petal_length[i] < split)
{
l1.push_back(species[i]);
}
else
{
l2.push_back(species[i]);
}
}
// Calculate information gain
double l1_entropy = calEntropy(l1);
double l2_entropy = calEntropy(l2);
double p1 = (double)l1.size()/s;
double p2 = (double)l2.size()/s;
information_gain = max(information_gain, all_entropy - l1_entropy*p1 - l2_entropy*p2);
}
return information_gain;
}
};
```

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