for uses iter(song) to loop; you can do this in your own code and then advance the iterator inside the loop; calling iter() on the iterable again will only return the same iterable object so you can advance the iterable inside the loop with for following right along in the next iteration.
Advance the iterator with the next() function; it works correctly in both Python 2 and 3 without having to adjust syntax:
song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
song_iter = iter(song)
for sing in song_iter:
print sing
if sing == 'look':
next(song_iter)
next(song_iter)
next(song_iter)
print 'a' + next(song_iter)
By moving the print sing line up we can avoid repeating ourselves too.
Using next() this way can raise a StopIteration exception, if the iterable is out of values.
You could catch that exception, but it'd be easier to give next() a second argument, a default value to ignore the exception and return the default instead:
song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
song_iter = iter(song)
for sing in song_iter:
print sing
if sing == 'look':
next(song_iter, None)
next(song_iter, None)
next(song_iter, None)
print 'a' + next(song_iter, '')
I'd use itertools.islice() to skip 3 elements instead; saves repeated next() calls:
from itertools import islice
song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
song_iter = iter(song)
for sing in song_iter:
print sing
if sing == 'look':
print 'a' + next(islice(song_iter, 3, 4), '')
The islice(song_iter, 3, 4) iterable will skip 3 elements, then return the 4th, then be done. Calling next() on that object thus retrieves the 4th element from song_iter().
Demo:
>>> from itertools import islice
>>> song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
>>> song_iter = iter(song)
>>> for sing in song_iter:
... print sing
... if sing == 'look':
... print 'a' + next(islice(song_iter, 3, 4), '')
...
always
look
aside
of
life
Hi all,
I am trying to skip values with an if statement in a for loop in Python and I cannot seem to do it.
What I am trying to do in Python is the following in C++:
[in]:
#include <iostream>
using namespace std;
int main()
{
for (int i = 0;i < 10;i++)
{if (i == 3) i++;
cout<<i;}
}
[out]: 012456789.
When I do this in Python, the for loop does not "skip" value 3, it just prints 4 twice instead of 3 and 4.
I can take this to an extreme and say if i=3: i == 20 and weirdly Python distinguishes between the i inside the for loop and the i with the given value 20 which baffles me.
[in]:
for i in range(10):if i == 3: i = i+1 print(i)
[out] :0 1 2 4 4 5 6 7 8 9
Is there any solution for this?
Thank you in advance for the help!
Python skip next index in loop without next(iterator) possible?
python - Skip multiple iterations in loop - Stack Overflow
python - How do I skip a few iterations in a for loop - Stack Overflow
How can you skip ahead in a for loop?
Videos
for uses iter(song) to loop; you can do this in your own code and then advance the iterator inside the loop; calling iter() on the iterable again will only return the same iterable object so you can advance the iterable inside the loop with for following right along in the next iteration.
Advance the iterator with the next() function; it works correctly in both Python 2 and 3 without having to adjust syntax:
song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
song_iter = iter(song)
for sing in song_iter:
print sing
if sing == 'look':
next(song_iter)
next(song_iter)
next(song_iter)
print 'a' + next(song_iter)
By moving the print sing line up we can avoid repeating ourselves too.
Using next() this way can raise a StopIteration exception, if the iterable is out of values.
You could catch that exception, but it'd be easier to give next() a second argument, a default value to ignore the exception and return the default instead:
song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
song_iter = iter(song)
for sing in song_iter:
print sing
if sing == 'look':
next(song_iter, None)
next(song_iter, None)
next(song_iter, None)
print 'a' + next(song_iter, '')
I'd use itertools.islice() to skip 3 elements instead; saves repeated next() calls:
from itertools import islice
song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
song_iter = iter(song)
for sing in song_iter:
print sing
if sing == 'look':
print 'a' + next(islice(song_iter, 3, 4), '')
The islice(song_iter, 3, 4) iterable will skip 3 elements, then return the 4th, then be done. Calling next() on that object thus retrieves the 4th element from song_iter().
Demo:
>>> from itertools import islice
>>> song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
>>> song_iter = iter(song)
>>> for sing in song_iter:
... print sing
... if sing == 'look':
... print 'a' + next(islice(song_iter, 3, 4), '')
...
always
look
aside
of
life
>>> song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
>>> count = 0
>>> while count < (len(song)):
if song[count] == "look" :
print song[count]
count += 4
song[count] = 'a' + song[count]
continue
print song[count]
count += 1
Output:
always
look
aside
of
life
You cannot alter the target list (i in this case) of a for loop. Use a while loop instead:
while i < 10:
i += 1
if i == 2:
i += 3
Alternatively, use an iterable and increment that:
from itertools import islice
numbers = iter(range(10))
for i in numbers:
if i == 2:
next(islice(numbers, 3, 3), None) # consume 3
By assigning the result of iter() to a local variable, we can advance the loop sequence inside the loop using standard iteration tools (next(), or here, a shortened version of the itertools consume recipe). for normally calls iter() for us when looping over a iterator.
The best way is to assign the iterator a name - it is common have an iterable as opposed to an iterator (the difference being an iterable - for example a list - starts from the beginning each time you iterate over it). In this case, just use the iter() built-in function:
numbers = iter(range(100))
Then you can advance it inside the loop using the name. The best way to do this is with the itertools consume() recipe - as it is fast (it uses itertools functions to ensure the iteration happens in low-level code, making the process of consuming the values very fast, and avoids using up memory by storing the consumed values):
from itertools import islice
import collections
def consume(iterator, n):
"Advance the iterator n-steps ahead. If n is none, consume entirely."
# Use functions that consume iterators at C speed.
if n is None:
# feed the entire iterator into a zero-length deque
collections.deque(iterator, maxlen=0)
else:
# advance to the empty slice starting at position n
next(islice(iterator, n, n), None)
By doing this, you can do something like:
numbers = iter(range(100))
for i in numbers:
...
if some_check(i):
consume(numbers, 3) # Skip 3 ahead.
I'm writing some code and when a specific condition is met inside of the for loop I want the index to be advanced. However, the following code doesn't seem to be doing that.
for i in range(0, n-2):
print (i)
if condition == True:
i = i + 3I ran it and it just prints all the values of i in order. I also added a print statement in the if condition to make sure that it was evaluating to true.
Why can't I skip ahead? Is that something that Python just doesn't support?