Split a string of numbers into lists of even and odd integers

Question

Having such a code

numbers = '1 2 3 4 5 6 7 8' nums = {'evens': [], 'odds': []} for number in numbers.split(' '): if int(number) % 2: nums['odds'].append(number) else: nums['evens'].append(number) 

How to do this on fewer lines?

Answer 1

Short code is not the best code. Short code is not fast code. Short code does not support code. Now, at the same time, it’s good to make your individual components concise and simple.

Here is what I will do:

 def split_odd_even(number_list): return { 'odds': filter(lambda n: (n % 2) != 0, number_list), 'evens': filter(lambda n: (n % 2) == 0, number_list) } def string_to_ints(string): return map(int, numbers.strip().split()) numbers = '1 2 3 4 5 6 7 8 9 10' nums = split_odd_even(string_to_ints(numbers)) print nums 

This gives me:

 {'odds': [1, 3, 5, 7, 9], 'evens': [2, 4, 6, 8, 10]} 

Although this code actually added a few lines in length, it became much clearer what the program does, since we applied Abstraction , and each component of the code does only one thing.

Despite the fact that we added two functions, the most noticeable part of the code left this:

 numbers = '1 2 3 4 5 6 7 8' nums = {'evens': [], 'odds': []} for number in numbers.split(' '): if int(number) % 2: nums['odds'].append(number) else: nums['evens'].append(number) 

For this:

 numbers = '1 2 3 4 5 6 7 8 9 10' nums = split_odd_even(string_to_ints(numbers)) 

And just reading these two lines, we know that numbers converted from a string to an int s list, and then we break these numbers into odd and even numbers and assign the result to nums .

To explain a couple of things that may not be familiar to everyone:

• map() calls a function for each element in list (or tuple or another iterable) and returns a new list with the result of the function called for each element. In this case, we use it to call int() for each item in the list.

• filter() calls a function for each element in list (or tuple or another iterable), which returns True or False for each element (well, veracity or false ) and returns a list of elements that evaluate to True when the function is called.

• Lambda expressions ( lambda ) are like "mini-functions" that take arguments and can be created in place.

Answer 2

Functional aproach:

 >>> numbers = '1 2 3 4 5 6 7 8' >>> numbers = map(int, numbers.split()) >>> nums = {'evens': filter(lambda x: x%2 == 0, numbers), 'odds': filter(lambda x: x%2 != 0, numbers)} >>> nums {'evens': [2, 4, 6, 8], 'odds': [1, 3, 5, 7]} 

Answer 3

You can accomplish the same results with itertools.groupby , for example:

 >>> from itertools import groupby >>> >>> numbers = '1 2 3 4 5 6 7 8' >>> d = {'even':[], 'odd':[]} >>> mynum = [int(x) for x in numbers.strip().split()] >>> for k,g in groupby(mynum, lambda x: x % 2): if k: d['odd'].extend(g) else: d['even'].extend(g) >>> d {'even': [2, 4, 6, 8], 'odd': [1, 3, 5, 7]} 

Answer 4

 numbers = '1 2 3 4 5 6 7 8' nums = {} nums["even"] = [int(i) for i in numbers.split() if int(i) % 2 == 0] nums["odd"] = [int(i) for i in numbers.split() if int(i) % 2 == 1] print(nums) 

Output:

 {'even': [2, 4, 6, 8], 'odd': [1, 3, 5, 7]} 

Answer 5

If you just want to try:

 numbers = '1 2 3 4 5 6 7 8' nums = {'evens': [], 'odds': []} for number in numbers.split(' '): category = 'odds' if int(number) % 2 else 'evens' nums[category].append(number) 

But if you want to use it in production: Readable code is more important than short code

Answer 6

You can do this as a single line, but I would not recommend it. Your code is ok.

 [nums['odds'].append(n) if int(n)%2 else nums['evens'].append(n) for n in numbers.split(' ')]