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From: "Walter =?utf-8?q?D=C3=B6rwald?=" <walter@livinglogic.de>
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Subject: Re: [Python-de] strings zusammensetzen.
Date: Tue, 29 Aug 2017 17:21:15 +0200
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On 26 Aug 2017, at 13:29, Peter Otten wrote:

> Tobias Herp wrote:
>
>>> d =3D "{base}{revision}{suffix}".format(base =3D a,
>>> revision =3D b,
>>> suffix =3D c)
>>
>> Wenn's denn unbedingt ein Template sein soll, w=C3=BCrde das bei mir=20
>> meistens
>> wie folgt aussehen:
>>
>> d =3D '%(base)s%(revision)s%(suffix)s' % locals()
>>
>
> Wurde schon erw=C3=A4hnt, dass ab 3.6 alternativ dazu auch
>
>>>> base, revision, suffix =3D "foo", "bar", "baz"
>>>> f"{base}{revision}{suffix}"
> 'foobarbaz'
>
> m=C3=B6glich ist?
>
> Das vermeidet ggf. das Erstellen eines tempor=C3=A4ren dicts.

Das ist auch mit Abstand die schnellste Variante:

Python 3.6.2 (default, Jul 26 2017, 16:42:24)
Type 'copyright', 'credits' or 'license' for more information
IPython 6.1.0 -- An enhanced Interactive Python. Type '?' for help.
In [1] =E2=96=B6 base, revision, suffix =3D "foo", "bar", "baz"
In [2] =E2=96=B6 %timeit base + revision + suffix
208 ns =C2=B1 0.532 ns per loop (mean =C2=B1 std. dev. of 7 runs, 1000000=
 loops=20
each)
In [3] =E2=96=B6 %timeit '%s%s%s' % (base, revision, suffix)
315 ns =C2=B1 1.55 ns per loop (mean =C2=B1 std. dev. of 7 runs, 1000000 =
loops=20
each)
In [4] =E2=96=B6 %timeit '{}{}{}'.format(base, revision, suffix)
462 ns =C2=B1 1.23 ns per loop (mean =C2=B1 std. dev. of 7 runs, 1000000 =
loops=20
each)
In [5] =E2=96=B6 %timeit f'{base}{revision}{suffix}'
14.6 ns =C2=B1 0.00911 ns per loop (mean =C2=B1 std. dev. of 7 runs, 1000=
00000=20
loops each)

Servus,
    Walter