data truncate when str(time.time()) in Python

Question

Can anyone tell me what happened when str(time.time()) runs? It seems the data has been truncated, but it recovered when float(str(time.time())) runs. I'm so confused.

In [1]: import time

In [2]: a = time.time()

In [3]: a
Out[3]: 1445007755.321532

In [4]: str(a)
Out[4]: '1445007755.32'

In [6]: a.__str__()
Out[6]: '1445007755.32'

In [7]: float(a)
Out[7]: 1445007755.321532

Check http://stackoverflow.com/q/1317558/1860929 – Anshul Goyal Oct 16 '15 at 15:07 — Anshul Goyal, Oct 16 '15 at 15:07
str truncate the value after the "." to be more readable. – Alexandre Mazel Oct 16 '15 at 15:49 — Alexandre Mazel, Oct 16 '15 at 15:49

Bakuriu · Answer 1 · 2015-10-16T15:13:36.743

Calling str(a) does not modify a, so there is no reason why float(a) should depend on str(a). Maybe you wanted to reassign a?

a = 1445007755.321532
a = str(a)
float(a)

In this case the result is different:

>>> a = 1445007755.321532
>>> b = str(a)
>>> a == float(b)
False
>>> a, b, float(b)
(1445007755.321532, '1445007755.32', 1445007755.32)

Note that it holds that float(x) == x if x is already a float, so what you see is not surprising at all.

^{Even more: float(x) is x if x is of type float.}

If you wonder why str seems to truncate the number it is because it returns a "human readable" representation of the value as a string.

If you want an "exact" representation use repr, or simply use a proper formatting function like: "%.6f" % a to control the number of decimal digits etc.

thanks, I also find a post here, http://stackoverflow.com/questions/1436703/difference-between-str-and-repr-in-python , which will be valuable for this answer, thanks a again. @Bakuriu — taotao.li, Oct 16 '15 at 15:17

score 0 · Answer 2 · answered Oct 16 '15 at 17:03

The difference in the number of digits after the decimal point is the difference between str(f) and repr(f).

By default, the values are converted to string (before displaying in the ipython console), using repr():

In [1]: class C:
   ...:     def __str__(self):
   ...:         return 'str'
   ...:     def __repr__(self):
   ...:         return 'repr'
   ...:     

In [2]: C()
Out[2]: repr

In [3]: str(C())
Out[3]: 'str'

btw, I can't reproduce your output on Python 3.4:

In [4]: 1445007755.321532
Out[4]: 1445007755.321532

In [5]: str(1445007755.321532)
Out[5]: '1445007755.321532'

In [6]: 1445007755.321532 .__str__()
Out[6]: '1445007755.321532'

But I can reproduce it on Python 2:

In [1]: 1445007755.321532
Out[1]: 1445007755.321532

In [2]: str(1445007755.321532)
Out[2]: '1445007755.32'

In [3]: 1445007755.321532 .__str__()
Out[3]: '1445007755.32'

Note: float() does NOT restore the precision here:

In [4]: float('1445007755.32')
Out[4]: 1445007755.32

In [5]: float(1445007755.32)
Out[5]: 1445007755.32

float(a) shows more digits in your question because a is already a float (it is probably a noop operation because floats are immutable):

In [6]: a = 1445007755.321532

In [7]: a is float(a)
Out[7]: True

i.e., float(a) may return the exact same object in this case.

data truncate when str(time.time()) in Python

2 Answers2