In my experience, at least for lists of ints, using enumerate just to find one index and throwing all other indexes away is so wasteful that it's faster to first only find the element and then use list.index to find its index. Based on some testing, I expect it to be about factor 1.44 faster than the explicit_loop solution (which uses enumerate) in @enke's benchmark.

def start_over_18(p):
    for x in p:
        if x > 18:
            return p[p.index(x):]
    return []

Another solution, using an iterator so I don't have to bother with indexes at all. Seems to be about twice as fast as the explicit_loop solution:

def start_over_18(p):
    it = iter(p)
    for x in it:
        if x > 18:
            return [x, *it]
    return []

I noticed the OP mention under an answer that p is actually a Pandas DataFrame. Here is a method of filtering all elements up to the first instance of a number greater than 18 using Pandas:

import pandas as pd
df = pd.DataFrame([4,9,10,4,20,13,29,3,39])
df = df[df[0].gt(18).idxmax():]
print(df)

Outputs:

    0
4  20
5  13
6  29
7   3
8  39

Note: I'm blind to the actual structure of your DataFrame so I just used exactly what was given.

You can use itertools.dropwhile:

from itertools import dropwhile

p = [4,9,10,4,20,13,29,3,39]

p = dropwhile(lambda x: x <= 18, p)
print(*p) # 20 13 29 3 39

In my opinion, this is arguably the easiest-to-read version. This also corresponds to a common pattern in other functional programming languages, such as dropWhile (<=18) p in Haskell and p.dropWhile(_ <= 18) in Scala.

---

Alternatively, using walrus operator (only available in python 3.8+):

exceeded = False
p = [x for x in p if (exceeded := exceeded or x > 18)]
print(p) # [20, 13, 29, 3, 39]

But my guess is that some people don't like this style. In that case, one can do an explicit for loop (ilkkachu's suggestion):

for i, x in enumerate(p):
    if x > 18:
        output = p[i:]
        break
else:
    output = [] # alternatively just put output = [] before for

Great solutions here, just wanted to demonstrate how to do it with numpy:

>>> import numpy as np
>>> p[(np.array(p) > 18).argmax():]
[20, 13, 29, 3, 39]

---

Since there are a lot of nice answers here, I decided to run some simple benchmarks. The first one uses the OP's sample array ([4,9,10,4,20,13,29,3,39]) of length 9. The second uses randomly generated array of length 20 thousand, where the first half is between 0 and 15, and the second half is between -20 and 30 (so that the split wouldn't occur right in the center).

Using the OP's data (array of length 9):

%timeit enke()
650 ns ± 15.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%timeit j1lee1()
546 ns ± 4.22 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%timeit j1lee2()
551 ns ± 19 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%timeit j2lee3()
536 ns ± 12.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%timeit richardec()
2.08 µs ± 16 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

Using an array of length 20,000 (20 thousand):

%timeit enke()
1.5 ms ± 34.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit j1lee1()
1.95 ms ± 43 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit j1lee2()
2.1 ms ± 53.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit j2lee3()
2.33 ms ± 96.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit richardec()
13.3 µs ± 461 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

Code to generate second array:

p = np.hstack([np.random.randint(0,15,10000),np.random.randint(-20,30,10000)])

So, for the small case, numpy is a slug and not needed. But the large case, numpy is almost 100x times faster and the way to go! :)

should not break suffice?

something like:

import numpy as np


input= np.array([4,9,10,4,20,13,29,3,39])
for index, item in enumerate(input):
  if(item>=18):
    print(input[index:]) 
    break

more info at:

https://www.poftut.com/control-python-loop-break-statement/#:~:text=Python%20provides%20for%20loops%20in,according%20to%20the%20current%20condition.

https://problemsolvingwithpython.com/05-NumPy-and-Arrays/05.06-Array-Slicing/