{"id":3906,"date":"2021-10-31T17:40:08","date_gmt":"2021-10-31T20:40:08","guid":{"rendered":"https:\/\/renanmf.com\/?p=3906"},"modified":"2023-04-16T15:11:30","modified_gmt":"2023-04-16T18:11:30","slug":"python-for-loop-increment-by-2","status":"publish","type":"post","link":"https:\/\/renanmf.com\/python-for-loop-increment-by-2\/","title":{"rendered":"Python for loop increment by 2"},"content":{"rendered":"

A regular for<\/code> loop will increase its iteration counter by one in each iteration.<\/p>\n

But there are situations where you want to increment the iteration counter by 2.<\/p>\n

For some reason you might want to work with only even values.<\/p>\n

Let’s see a few solutions for this.<\/p>\n

range function<\/h2>\n

The solution to increment a for loop by 2 in Python is to use the range()<\/code> function.<\/p>\n

This function allows you to specify three parameters: start<\/code>, stop<\/code>, and step<\/code>.<\/p>\n

start<\/code> is the counter’s initial value, step<\/code> is by how much you want to increment until you reach the value of stop - 1<\/code>, because the value of stop is included.<\/p>\n

The example bellow shows how to use range()<\/code> with a step of 2 starting from 0.<\/p>\n

for i in range(0,20,2):\n    print(i)<\/code><\/pre>\n
The output:<\/p>\n
0\n2\n4\n6\n8\n10\n12\n14\n16\n18<\/code><\/pre>\n
Of course, you can use range()<\/code> to specify any step you would like, so if you want a step of 5, just do:<\/p>\n
for i in range(0,20,5):\n    print(i)<\/code><\/pre>\n
The output is:<\/p>\n
0\n5\n10\n15<\/code><\/pre>\n
Slicing<\/h2>\n
If you never heard of slicing before, I recommend you to read Understanding Slicing in Python<\/a> first, there is also a video linked in the article if you prefer.<\/p>\n
Slicing is useful when you want to apply a different step when working with a pre-defined list.<\/p>\n
In this case I have a list numbers<\/code> with the number from 1 to 16.<\/p>\n
The logic is very similar to range()<\/code>, since you also have a start<\/code>, a stop<\/code>, and a step<\/code>.<\/p>\n
In this case I’m starting on index 1, which is the number 2 in the list, remember lists are 0-indexed in Python.<\/p>\n
I won’t put any stop<\/code> value since I want to go until the last index.<\/p>\n
Finally, I put a step of 2.<\/p>\n
numbers = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]\n\nfor i in numbers[1::2]:\n    print(i)<\/code><\/pre>\n
The output will be:<\/p>\n
2\n4\n6\n8\n10\n12\n14\n16<\/code><\/pre>\n
Another way to implement the slicing from the code above is to combine range()<\/code> with len()<\/code><\/a>:<\/p>\n
numbers = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]\n\nfor i in range(1, len(numbers), 2):\n   print(numbers[i])<\/code><\/pre>\n
The output is the same:<\/p>\n
2\n4\n6\n8\n10\n12\n14\n16<\/code><\/pre>\n
Again, the same way we could use any step in range()<\/code>, we can change the step in slicing to anything we would like.<\/p>\n
What about about float values?<\/h2>\n
An extra trick is how to achieve the same thing with float values.<\/p>\n
In this case you have to use the library numpy and its function arange<\/code>.<\/p>\n
This way you can use floats as start<\/code>, stop<\/code>, and step<\/code>.<\/p>\n
import numpy as np\n\nfor i in np.arange(1.5, 2.5, 0.1):\n    print (i)<\/code><\/pre>\n
The output is:<\/p>\n
1.5\n1.6\n1.7000000000000002\n1.8000000000000003\n1.9000000000000004\n2.0000000000000004\n2.1000000000000005\n2.2000000000000006\n2.3000000000000007\n2.400000000000001<\/code><\/pre>\n
Watch this on Youtube:<\/h2>\n