如何在 Python 中按顺序获取列表元素的所有串联？

Question

background

I have lists of numbers and symbols like:

[1,':',2022,'.','01','.','01']

I would like to call a function like is_date(string) on every concatenation of the elements in order.

---

minimal example

Considering the simple example [1,2,3] as input, the expected list of concatenations should be: ['1','12','2','123','23','3']

I would like not to get '13' in the results

So afterwards I could call:

for token in list_of_concatenation:
    if is_date(token):
        do something

Answer 1

You can use a double loop, slicing, and string concatenation:

l = [1, 2, 3]
l2 = [str(x) for x in l]
[''.join(l2[i:j+1]) for i in range(len(l2)) for j in range(i, len(l2))]

output: ['1', '12', '123', '2', '23', '3']

Answer 2

How about a nested for loop:

sample = [1,':',2022,'.','01','.','01']
sample_str = [str(s) for s in sample]

for ix, i in enumerate(sample_str):
    for j in range(ix, len(sample_str)):
        print(''.join(sample_str[ix:j+1]))

output:

1
1:
1:2022
1:2022.
1:2022.01
1:2022.01.
1:2022.01.01
:
:2022
:2022.
:2022.01
:2022.01.
:2022.01.01
2022
2022.
2022.01
2022.01.
2022.01.01
.
.01
.01.
.01.01
01
01.
01.01
.
.01
01

Answer 3

You are not looking for the itertools.powerset() recipe because it ignores the "order":

from itertools import  chain, combinations

# source:  https://docs.python.org/3/library/itertools.html
def powerset(iterable):
    "powerset([1,2,3]) --> () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)"
    s = list(iterable)

    # use  range(1, len(s)+1) to avoid empty result -> comment @mozway
    return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))

d = [1,2,3]

print(list(powerset(d)))  

wich returns tuples as iterator (that I listified):

[(), (1,), (2,), (3,), (1, 2), (1, 3), (2, 3), (1, 2, 3)]

You can filter the result to remove the empty tuple and convert tuples to strings using

result = [''.join(map(str, tup)) for tup in powerset(d) if tup]

print(result)

to get

['1', '2', '3', '12', '13', '23', '123']

• for a not that efficient solution you could use that to filter down to what you really really want.