在python中生成数字和交替字符串的序列

Question

Aim

I would like to generate a sequence as list in python , such as:

['s1a', 's1b', 's2a', 's2b', ..., 's10a', 's10b']

Properties:

• items contain a single prefix
• numbers are sorted numerical
• suffix is alternating per number

Approach

To get this, I applied the following code, using an xrange and comprehensive list approach:

# prefix
p = 's'
# suffix
s = ['a', 'b']
# numbers
n = [ i + 1 for i in list(xrange(10))]
# result
[ p + str(i) + j for i, j in zip(sorted(n * len(s)), s * len(n)) ]

Question

Is there a more simple syntax to obtain the results, eg using itertools ? Similar to this question?

Answer 1

双重理解列表可以实现以下目的：

['s'+str(x)+y for x in range(1,11) for y in 'ab']

Answer 2

itertools.product might be your friend:

all_combos = ["".join(map(str, x)) for x in itertools.product(p, n, s)]

returns:

['s1a', 's1b', 's2a', 's2b', 's3a', 's3b', 's4a', 's4b', 's5a', 's5b', 's6a', 's6b', 's7a', 's7b', 's8a', 's8b', 's9a', 's9b', 's10a', 's10b']

EDIT: as a one-liner:

all_combos = ["".join(map(str,x)) for x in itertools.product(['s'], range(1, 11), ['a', 'b'])]

EDIT 2: as pointed out in James' answer, we can change our listed string element in the product call to just strings, and itertools will still be able to iterate over them, selecting characters from each:

all_combos = ["".join(map(str,x)) for x in itertools.product('s', range(1, 11), 'ab')]

Answer 3

How about:

def func(prefix,suffixes,size):
    k = len(suffixes)
    return [prefix+str(n/k+1)+suffixes[n%k] for n in range(size*k)]

# usage example:
print func('s',['a','b'],10)

This way you can alternate as many suffixes as you want.

And of course, each one of the suffixes can be as long as you want.

Answer 4

You can use a double-list comprehension, where you iterate on number and suffix. You don't need to load any

Below is a lambda function that takes 3 parameters, a prefix, a number of iterations, and a list of suffixes

foo = lambda prefix,n,suffix: list(prefix+str(i)+s for s in suffix for i in range(n))

You can use it like this

foo('p',10,'abc')

Or like that, if your suffixes have more than one letter

foo('p',10,('a','bc','de'))

Answer 5

For maximum versatility I would do this as a generator. That way you can either create a list, or just produce the sequence items as they are needed.

Here's code that runs on Python 2 or Python 3.

def psrange(prefix, suffix, high):
    return ('%s%d%s' % (prefix, i, s) for i in range(1, 1 + high) for s in suffix)

res = list(psrange('s', ('a', 'b'), 10))
print(res)

for s in psrange('x', 'abc', 3):
    print(s)

output

['s1a', 's1b', 's2a', 's2b', 's3a', 's3b', 's4a', 's4b', 's5a', 's5b', 's6a', 's6b', 's7a', 's7b', 's8a', 's8b', 's9a', 's9b', 's10a', 's10b']
x1a
x1b
x1c
x2a
x2b
x2c
x3a
x3b
x3c