[英]Printing a tuple in Python with user-defined precision
Following Printing tuple with string formatting in Python , I'd like to print the following tuple: 在Python中使用字符串格式打印元组之后,我想打印以下元组:
tup = (0.0039024390243902443, 0.3902439024390244, -0.005853658536585366, -0.5853658536585366)
with only 5 digits of precision. 仅有5位数的精度。 How can I achieve this?
我该如何实现?
(I've tried print("%.5f" % (tup,))
but I get a TypeError: not all arguments converted during string formatting
). (我尝试过
print("%.5f" % (tup,))
但出现TypeError: not all arguments converted during string formatting
)。
You can print the floats with custom precision "like a tuple": 您可以“像元组”一样以自定义精度打印浮点数:
>>> tup = (0.0039024390243902443, 0.3902439024390244, -0.005853658536585366, -0.5853658536585366)
>>> print('(' + ', '.join(('%.5f' % f) for f in tup) + ')')
(0.00390, 0.39024, -0.00585, -0.58537)
尝试以下方法(列表理解)
['%.5f'% t for t in tup]
you can work on single item. 您可以处理单个项目。 Try this:
尝试这个:
>>> tup = (0.0039024390243902443, 0.3902439024390244, -0.005853658536585366, -0.5853658536585366)
>>> for t in tup:
print ("%.5f" %(t))
0.00390
0.39024
-0.00585
-0.58537
You can iterate over tuple like this, and than you can print result for python > 3 您可以像这样遍历元组,然后可以打印python> 3的结果
["{:.5f}".format(i) for i in tup]
And for python 2.7 对于python 2.7
['%.5f'% t for t in tup]
Possible workaround: 可能的解决方法:
tup = (0.0039024390243902443, 0.3902439024390244, -
0.005853658536585366, -0.5853658536585366)
print [float("{0:.5f}".format(v)) for v in tup]
Most Pythonic way to achieve this is with map()
and lambda()
function. 实现此目标的大多数Python方法是使用
map()
和lambda()
函数。
>>> map(lambda x: "%.5f" % x, tup)
['0.00390', '0.39024', '-0.00585', '-0.58537']
I figured out another workaround using Numpy: 我想出了另一个使用Numpy的解决方法:
import numpy as np
np.set_printoptions(precision=5)
print(np.array(tup))
which yields the following output: 产生以下输出:
[ 0.0039 0.39024 -0.00585 -0.58537]
Here's a convenient function for python >3.6 to handle everything for you: 这是python> 3.6的一个便捷函数,可以为您处理所有事情:
def tuple_float_to_str(t, precision=4, sep=', '):
return '({})'.format(sep.join(f'{x:.{precision}f}' for x in t))
Usage: 用法:
>>> print(funcs.tuple_float_to_str((12.3456789, 8), precision=4))
(12.3457, 8.0000)
Try this: 尝试这个:
class showlikethis(float):
def __repr__(self):
return "%0.5f" % self
tup = (0.0039024390243902443, 0.3902439024390244, -0.005853658536585366, -0.5853658536585366)
tup = map(showlikethis, tup)
print tup
You may like to re-quote your question, tuple dnt have precision. 您可能想重新引用您的问题,元组dnt具有精度。
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