Finding all possible combination of alphabets of length 5

Question

The code finds all the combination of alphabet with length 5 and saves it in a list(its ASCII value, minus 97). I want to optimize the code to reduce execution time.

from itertools import product
def permuteKey():

    def pp(chunks):
        return map(''.join, chunks)

    k=pp(product('abcdefghijklmnopqrstuvwxyz', repeat=5))
    key=[]
    for i in range(len(k)):
        temp=k[i]
        temp2=[]
        for j in range(len(temp)):
            temp1=temp[j]
            temp2.append(ord(temp1)-97)
        key.append(temp2)
        print key

permuteKey()

Answer 1

Let me take a stab at this.

from itertools import product
def permuteKey():

    def pp(chunks):
        return map(''.join, chunks)

    k=pp(product('abcdefghijklmnopqrstuvwxyz', repeat=2))
    key=[]
    for i in range(len(k)):
        temp=k[i]
        temp2=[]
        for j in range(len(temp)):
            temp1=temp[j]
            temp2.append(ord(temp1)-97)
        key.append(temp2)
        print key

permuteKey()

Let's start by taking a look at the inner loop:

    temp=k[i]
    temp2=[]
    for j in range(len(temp)):
        temp1=temp[j]
        temp2.append(ord(temp1)-97)

First of all, you're only referring to temp[j] once, and never use j any further. The quick solution would be to iterato over temp instead:

    temp=k[i]
    temp2=[]
    for temp1 in temp:
        temp2.append(ord(temp1)-97)

But now we easily see that temp2 can be written as a list comprehension:

    temp=k[i]
    temp2=[ord(temp1) - 97 for temp1 in temp]

The temp = k[i] can also be removed, leaving

    temp2=[ord(temp1) - 97 for temp1 in k[i]]

Now we end up with

from itertools import product
def permuteKey():

    def pp(chunks):
        return map(''.join, chunks)

    k=pp(product('abcdefghijklmnopqrstuvwxyz', repeat=2))
    key=[]
    for i in range(len(k)):
        temp2=[ord(temp1) - 97 for temp1 in k[i]]
        key.append(temp2)
        print key

permuteKey()

Again, we can apply the trick that k[i] is only used once:

from itertools import product
def permuteKey():

    def pp(chunks):
        return map(''.join, chunks)

    k=pp(product('abcdefghijklmnopqrstuvwxyz', repeat=2))
    key=[]
    for val in k:
        temp2=[ord(temp1) - 97 for temp1 in val]
        key.append(temp2)
        print key

permuteKey()

Another thing I notice, is that you never use the value of one of the items itself, but always ord(c) - 97. Why not just replace abcd...xyz with range(26), and remove the ord(...) - 97 part?

from itertools import product
def permuteKey():

    def pp(chunks):
        return map(''.join, chunks)

    k=product(range(26), repeat=2)
    key=[]
    for val in k:
        temp2=[temp1 for temp1 in val]
        key.append(temp2)
        print key

permuteKey()

But now we don't need pp anymore.

from itertools import product
def permuteKey():

    k=product(range(26), repeat=2)
    key=[]
    for val in k:
        temp2=[temp1 for temp1 in val]
        key.append(temp2)
        print key

permuteKey()

Also, the definition of temp2 is now equal to val.

from itertools import product
def permuteKey():

    k=product(range(26), repeat=2)
    key=[]
    for val in k:
        temp1 = list(val)  # `val` is a tuple, but your code expects a list.
        key.append(temp1)
        print key

permuteKey()

Does it really matter that you want lists instead of tuples? I think not.

from itertools import product
def permuteKey():

    k=product(range(26), repeat=2)
    key=[]
    for val in k:
        key.append(val)
        print key

permuteKey()

Now, I have to wonder: why are we actually printing key at every step? I think you might have made a typo, so I'm going to assume you meant the following: (If not, skip the review from here).

from itertools import product
def permuteKey():

    k=product(range(26), repeat=2)
    key=[]
    for val in k:
        key.append(val)
    print key

permuteKey()

Now, it's bad practise to print inside a function. Better would it be to return the value.

from itertools import product
def permuteKey():

    k=product(range(26), repeat=2)
    key=[]
    for val in k:
        key.append(val)
    return key

print permuteKey()

Now, notice the similarity between k and key. key = list(k).

from itertools import product
def permuteKey():

    k=product(range(26), repeat=2)
    return list(k)

print permuteKey()

But, now you have an intermediate value, you do not need it, and you can remove it.

from itertools import product
def permuteKey():
    return list(product(range(26, repeat=2)))

print permuteKey()

Now, I don't know how often you need permuteKey. If only once, you can write

from itertools import product
print list(product(range(26), repeat=2))