如何在不舍入的情况下将值修剪到点后的两个位置，python [英] How to trim value to two places after dot without rounding, python

问题描述

我正在寻找最快的方法来获取我的列中的每个值，而点后仅两位数字而不使用round()

I am searching for fastest method to get each value in my column with only two digits after dot without using round()

pd.Series:

输入:

1.42345
12.33444
111.66777
2.059999

预期输出:

1.42
12.33
111.66
2.05

我正在考虑将其转换为字符串，然后使用slice，但是在切片中，我需要使用开始和停止选项，当前面的数字长度不同时，它会比较困难.

I am thinking to convert it to string and then use slice but in slice i need to use start and stop options, its hard when digits before are in different lenght.

如果没有其他选择，我将使用.str.extract(r'([0-9]{1,5}.[0-9]{2})')，但是也许有吗?显示浮动而不舍入时有一些限制吗?

If there will be no other option I will use .str.extract(r'([0-9]{1,5}.[0-9]{2})') , but maybe there is? some limitations to show float without rounding?

它不是重复的主题，对此的解决方案是舍入值:截断为三位小数在Python中

its not duplicate topic, solution from this one is rounding values :Truncate to three decimals in Python

推荐答案

一种选择是采用

One option is to take the floordiv by 0.01 and to divide again the value by 100:

s.floordiv(0.01).div(100)

0      1.42
1     12.33
2    111.66
3      2.05
dtype: float64

---

显然，它比强制转换为字符串并提取至小数点后第二位要好:

---

It clearly performs better than casting to string and extracting up to the second decimal place:

s = pd.Series(np.random.randn(1_000_000))

%timeit s.astype(str).str.extract(r'(\d+\.\d{2})')
# 1.76 s ± 42.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit s.floordiv(0.01).div(100)
# 42.1 ms ± 3.08 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

%timeit s//0.01/100
# 40.5 ms ± 3.31 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

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