NLTK package to estimate the (unigram) perplexity

Perplexity is the inverse probability of the test set, normalized by the number of words. In the case of unigrams:

Now you say you have already constructed the unigram model, meaning, for each word you have the relevant probability. Then you only need to apply the formula. I assume you have a big dictionary unigram[word] that would provide the probability of each word in the corpus. You also need to have a test set. If your unigram model is not in the form of a dictionary, tell me what data structure you have used, so I could adapt it to my solution accordingly.

perplexity = 1
N = 0

for word in testset:
    if word in unigram:
        N += 1
        perplexity = perplexity * (1/unigram[word])
perplexity = pow(perplexity, 1/float(N))

UPDATE:

As you asked for a complete working example, here's a very simple one.

Suppose this is our corpus:

corpus ="""
Monty Python (sometimes known as The Pythons) were a British surreal comedy group who created the sketch comedy show Monty Python's Flying Circus,
that first aired on the BBC on October 5, 1969. Forty-five episodes were made over four series. The Python phenomenon developed from the television series
into something larger in scope and impact, spawning touring stage shows, films, numerous albums, several books, and a stage musical.
The group's influence on comedy has been compared to The Beatles' influence on music."""

Here's how we construct the unigram model first:

import collections, nltk
# we first tokenize the text corpus
tokens = nltk.word_tokenize(corpus)

#here you construct the unigram language model 
def unigram(tokens):    
    model = collections.defaultdict(lambda: 0.01)
    for f in tokens:
        try:
            model[f] += 1
        except KeyError:
            model [f] = 1
            continue
    N = float(sum(model.values()))
    for word in model:
        model[word] = model[word]/N
    return model

Our model here is smoothed. For words outside the scope of its knowledge, it assigns a low probability of 0.01. I already told you how to compute perplexity:

#computes perplexity of the unigram model on a testset  
def perplexity(testset, model):
    testset = testset.split()
    perplexity = 1
    N = 0
    for word in testset:
        N += 1
        perplexity = perplexity * (1/model[word])
    perplexity = pow(perplexity, 1/float(N)) 
    return perplexity

Now we can test this on two different test sets:

testset1 = "Monty"
testset2 = "abracadabra gobbledygook rubbish"

model = unigram(tokens)
print perplexity(testset1, model)
print perplexity(testset2, model)

for which you get the following result:

>>> 
49.09452736318415
99.99999999999997

Note that when dealing with perplexity, we try to reduce it. A language model that has less perplexity with regards to a certain test set is more desirable than one with a bigger perplexity. In the first test set, the word Monty was included in the unigram model, so the respective number for perplexity was also smaller.