How to skip the next two iterations in for loop using python?-CodePudding

In a for loop, let's assume it contains 100 iterations, and what I need is, i need to print the first element and skip the next two element, and then need to prit the fourth element.. for example,

0 #should print 
1 #should skip
2 #should skip
3 #should print 
4 #should skip
5 #should skip
6 #should print
.
. like wise

I tried some existing skipping solutions out this platform, but didn't fit for my problem. I know the continue statement allows you to skip over the current iteration, but I can't figure out how to skip the next two iterations. Ttried some itertool fuctions also.

CodePudding user response：

If you iterate on your own range, generate one over three elements

values = ["dog0", "dog1", "dog2", "dog3", "dog4", "dog5", "dog6"]
for i in range(0, 8, 3):
    print(i, values[i])

If you iterate over an iterable of values, use enumerate to filter

values = ["dog0", "dog1", "dog2", "dog3", "dog4", "dog5", "dog6"]
for i, value in enumerate(values):
    if i % 3 != 0:
        continue
    print(i, value)

CodePudding user response：

You can do something like this

arr = [1,2,3,4,5,6,7,8,9,10]
skip_count = 2  # how many to skip
counter = skip_count

for num in arr:
    if counter == skip_count:
        counter = 0
        print(num)
    else:
        counter  = 1

Alternatively, this will work too

arr = [1,2,3,4,5,6,7,8,9,10]

for index in range(0, len(arr), 3):
    print(arr[index])

CodePudding user response：

One possible approach would be using enumerate and mod (%).

nums = [0, 1, 2, 3, 4, 5, 6, 7, 8]

for index, num in enumerate(nums):
    print(num) if index % 3 == 0 else None

CodePudding user response：

If it's a range you can use the optional step argument range(start, stop, step)

for k in range(start, stop, 3): print(k)

for k in range(len(my_data), step=3): print(k)

If it's a list you can use list slicing where once again the optional arguments for slicing is start, stop and step!

for k in my_array[::3]: print(k)