Using Break in a List Comprehension

Using break in a list comprehension

even = [n for n in numbers[:None if 412 not in numbers else numbers.index(412)] if not n % 2] 

Just took F.J.'s code above and added a ternary to check if 412 is in the list. Still a 'one liner' and will work even if 412 is not in the list.

break inside list comprehension

use any() with a generator

filterList=['test','2','x','123','orange']
print ([i for i in datalist if any(j for j in filterList if j in i) ])

any stops the iterations when the first element is found

list comprehension with if and break

You can create a filtered new list based on your keyword search in a second list using any:

>>> [x for x in queries.split(',') if any(kw in x for kw in keywords)]
['watch movie online', ' movie poster', ' course of events', ' movie in this summer', ' python course', ' watch movie about training course']

The break is naturally in any since it will short-circuit on the first True and break out of the loop over keywords.

Then print the new list or sequence using '\n'.join(iterable) to get the effect of printing each element in the loop.

>>> print ('\n'.join(x for x in queries.split(',') if any(kw in x for kw in keywords)))
watch movie online
 movie poster
 course of events
 movie in this summer
 python course
 watch movie about training course

In that case, you can use a generator vs a list comprehension.

However, don't be afraid of just using a loop in Python; they are easy to understand and obvious in intent.

(And your can strip the space on the elements using .lstrip(x) if desired.)

list comprehensions with break

You can't really break a list comprehension's internal for loop, what you can do is avoid having to break it at all by using the next function to find the first occurrence of a matching value:

list1 = [4, 5, 6, 9, 10, 16, 21, 23, 25, 27]
list2 = [1, 3, 5, 7, 8, 11, 12, 13, 14, 15, 17, 20, 24, 26, 56]
list3 = [ next(j for j in list2 if j>i) for i in list1 ]

output:

print(list1)
print(list3)
[4, 5, 6, 9, 10, 16, 21, 23, 25, 27]
[5, 7, 7, 11, 11, 17, 24, 24, 26, 56]

If you are concerned about performance (since the list comprehension will be slower than the loops), you could use a bisecting search in list 2 to find the next higher value:

from bisect import bisect_left
list3 = [ list2[bisect_left(list2,i+1)] for i in list1 ]

This assumes that list2 is sorted in ascending order and that max(list2) > max(list1)

Using next instead of break in comprehension list

You could pass a default argument in next().And next() would only return only one element,so "".join([]) is unnecessary.

Code below:

key = next((key for key, value in substitution_dict.items() if word == key or word in value), None)

When the iterator is exhausted, it would return None.

Or if you really want to use it with ''.join, like:

key = "".join([next((key for key, value in substitution_dict.items() if word == key or word in value), "")])

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