英文:
Why f-strings are slow compared to % notation?
问题
f-strings比其他字符串格式化方式更快。最近，我进行了一些小的研究，结果让人吃惊 - 使用f-strings的小段代码比百分号符号表示法需要更多的时间。有人可以解释原因吗？
代码如下：
from timeit import repeat
setup = 'i = 1.34234324'
for code in ['"%f %f" % (i, i)', "f'{i} {i}'"] * 5:
t = min(repeat(code, setup, number=10000)) / 10000
print(f'{int(t*1e9):4} ns ', code)
打印结果是：
321 ns "%f %f" % (i, i)
782 ns f'{i} {i}'
322 ns "%f %f" % (i, i)
786 ns f'{i} {i}'
321 ns "%f %f" % (i, i)
779 ns f'{i} {i}'
320 ns "%f %f" % (i, i)
782 ns f'{i} {i}'
320 ns "%f %f" % (i, i)
780 ns f'{i} {i}'
我使用的是Python 3.10.7。
我期望f-strings的性能更好。
我也希望得到哪种浮点数替代字符串的方式最快的答案。
英文:
I have been thought that f-strings are faster than other ways of string formatting.
Recently I have done a small research. The result was a surprise -- small code with f-strings takes more time than percentage notation. Can someone explain the reason why?
The code is following:
from timeit import repeat
setup = 'i = 1.34234324'
for code in ["'%f %f' % (i, i)", "f'{i} {i}'"] * 5:
t = min(repeat(code, setup, number=10000)) / 10000
print(f'{int(t*1e9):4} ns ', code)
The print result is :
321 ns '%f %f' % (i, i)
782 ns f'{i} {i}'
322 ns '%f %f' % (i, i)
786 ns f'{i} {i}'
321 ns '%f %f' % (i, i)
779 ns f'{i} {i}'
320 ns '%f %f' % (i, i)
782 ns f'{i} {i}'
320 ns '%f %f' % (i, i)
780 ns f'{i} {i}'
I use Python 3.10.7
I expected that f-strings would have greater performance.
I also would like to get the answer which way of float substituting into a string is the fastes.
答案1
得分: 1
以下是代码的翻译部分：
# 第一个代码段
i = 1.645649846546786534646568684564688646546548686465416574786
a = f"{i}"
print(a)
# 输出
1.6456498465467866
# 第二个代码段
i = 1.645649846546786534646568684564688646546548686465416574786
b = "%f" % (i)
print(b)
# 输出
1.645650
# 用16位小数精度进行比较的修改后的代码
"%.16f %.16f" % (i, i)
# 使用 timeit 进行性能比较
import timeit
timeit.timeit('a=f"{i}"', setup='i = 1.645649846546786534646568684564688646546548686465416574786', number=100000)
0.04724510000005466
timeit.timeit('a="%.16f" % (i)', setup='i = 1.645649846546786534646568684564688646546548686465416574786', number=100000)
0.03985800000009476
# f-string 与 % 相当
英文:
Look at below two snippets of code.
it will be slower for floats since f-string
does not truncate the float
and guarantees some 16 digits of precision
in all other cases it will be much faster
i=1.645649846546786534646568684564688646546548686465416574786
a=f"{i}"
print(a)
1.6456498465467866
i=1.645649846546786534646568684564688646546548686465416574786
b="%f"%(i)
print(b)
1.645650
You may ammend the code to following for
proper comparison use %f with 16 decimal points precision
"'%.16f %.16f' % (i, i)"
import timeit
timeit.timeit('a=f"{i}"',setup='i= 1.645649846546786534646568684564688646546548686465416574786',number=100000)
0.04724510000005466
timeit.timeit('a="%.16f"%(i)',setup='i= 1.645649846546786534646568684564688646546548686465416574786',number=100000)
0.03985800000009476
### f-string is comparable %
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