Question

我有一个由浮动价值组成的清单,但过于详细。我知道,我们可以通过使用<代码>(>%.f”%变量)的操作者,如:

result = [359.70000000000005]
result = "%.2f" % result
result = [359.70]

我的问题是,我怎样才能把价值清单变成其四舍五入的等数,而不用使用一个炉子。我曾尝试过一些东西,但投下了:

list = [0.30000000000000004, 0.5, 0.20000000000000001]
list = "%.2f" % list
TypeError: not all arguments converted during string formatting

我怎么能提供这样一个清洁的清单:

list = [0.30, 0.5, 0.20]

Answer 1

<代码>%.2f>不退还clean浮动。该浮标有2个小数。

my_list = [0.30000000000000004, 0.5, 0.20000000000000001]
my_formatted_list = [  %.2f  % elem for elem in my_list ]

返回:

[ 0.30 ,  0.50 ,  0.20 ]

此外,不打上你可变的<代码> >list。这是为编制清单而保留的一个词。使用其他一些名称,例如my_list 。

如果您希望获得<代码>[0.30,0.5, 0.20](或至少是最接近的浮动),你可以尝试:

my_rounded_list = [ round(elem, 2) for elem in my_list ]

返回:

[0.29999999999999999, 0.5, 0.20000000000000001]

Answer 2

如果你真的想要一种无休止的解决办法,那么你可以使用 n及其。

import numpy as np
myList = list(np.around(np.array(myList),2))

Answer 3

You might want to look at Python s decimal module, which can make using floating point numbers and doing arithmetic with them a lot more intuitive. Here s a trivial example of one way of using it to "clean up" your list values:

>>> from decimal import *
>>> mylist = [0.30000000000000004, 0.5, 0.20000000000000001]
>>> getcontext().prec = 2
>>> ["%.2f" % e for e in mylist]
[ 0.30 ,  0.50 ,  0.20 ]
>>> [Decimal("%.2f" % e) for e in mylist]
[Decimal( 0.30 ), Decimal( 0.50 ), Decimal( 0.20 )]
>>> data = [float(Decimal("%.2f" % e)) for e in mylist]
>>> data
[0.3, 0.5, 0.2]

Answer 4

mylist = [0.30000000000000004, 0.5, 0.20000000000000001]
myRoundedList =  [round(x,2) for x in mylist] 
# [0.3, 0.5, 0.2]

Answer 5

可在https://docs.python.org/3/library/Functions.html#map” rel=“noreferer”>。以及。表示:

my_list = [0.2111111111, 0.5, 0.3777777777]
my_list_rounded = list(map(lambda x: round(x, ndigits=2), my_list))
my_list_rounded                                                                                                                                                                                                                 
Out[3]: [0.21, 0.5, 0.38]

或者,您也可通过 partial,从:

from functools import partial my_list = [0.2111111111, 0.5, 0.3777777777] round_2digits = partial(round, ndigits=2) my_list_rounded = list(map(round_2digits, my_list)) my_list_rounded Out[6]: [0.21, 0.5, 0.38]

Answer 6

Another option which doesn t require numpy is:

precision = 2  
myRoundedList = [int(elem*(10**precision)+delta)/(10.0**precision) for elem in myList]

# delta=0 for floor
# delta = 0.5 for round
# delta = 1 for ceil

Answer 7

www.un.org/Depts/DGACM/index_spanish.htm 最容易的方式:

进口 n

页: 1

Answer 8

Python3中最简单的方法是使用<代码>numpy.round(),以形成一个四舍五入的阵列和list(>)将其重新列入一个名单。不必在将清单发送至<代码>numpy.round()之前将清单插入一个阵列。

import numpy as np

my_list = [0.342514, 0.52341234, 0.21234354346]

rounded_list = list(np.round(my_list, 2))

产出:<代码>[0.34,0.52, 0.21]

如果你不想把 n作为依赖物,那么 e蒸炉解决办法就是最好的选择。

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