let's say i have a list of strings like this
L = ['5', '3', '4', '1', '2', '2 3 5', '2 4 8', '5 22 1 37', '5 22 1 22', '5 22 1 23', ....]
How can i sort this list so that i would have something like this:
L = ['1', '2', '3','4', '5', '2 3 5', '2 4 8', '5 22 1 22', ' 5 22 1 23', '5 22 1 37', ...]
basically i need to order the list based on the first different number between 2 strings
You could sort using a tuple:
L = ['5', '3', '4', '1', '2', '2 3 5', '2 4 8', '5 22 1 37', '5 22 1 22', '5 22 1 23']
result = sorted(L, key=lambda x: (len(x.split()),) + tuple(map(int, x.split())))
print(result)
Output
['1', '2', '3', '4', '5', '2 3 5', '2 4 8', '5 22 1 22', '5 22 1 23', '5 22 1 37']
The idea is to use as key a tuple where the first element is the amount of numbers in the string and the rest is the tuple of numbers. For example for '2 3 5' the key is (3, 2, 3, 5)
As suggested by @PM2Ring you could use a def function instead of a lambda:
def key(x):
numbers = tuple(map(int, x.split()))
return (len(numbers),) + numbers
def function instead of a lambda for the key function so you don't have to call x.split twice on each item.
lambda x: (len(y:=x.split()),) + tuple(map(int, y))). (Personally, I'd still pre-define the function, but there appear to be people who like the syntax that PEP-572 settled on.)A slightly different approach than @Daniel's one.
idx = sorted(range(len(L)), key=lambda i: int(''.join(L[i].split())))
L = [L[i] for i in idx]
output
['1',
'2',
'3',
'4',
'5',
'2 3 5',
'2 4 8',
'5 22 1 22',
'5 22 1 23',
'5 22 1 37']
'4 2'appear in your result, before or after'5'?