问题:如何在Python中将数字四舍五入为有效数字

我需要四舍五入才能在UI中显示。例如,一个重要的数字:

1234-> 1000

0.12-> 0.1

0.012-> 0.01

0.062-> 0.06

6253-> 6000

1999-> 2000

是否有使用Python库执行此操作的好方法,还是我必须自己编写它?

I need to round a float to be displayed in a UI. E.g, to one significant figure:

1234 -> 1000

0.12 -> 0.1

0.012 -> 0.01

0.062 -> 0.06

6253 -> 6000

1999 -> 2000

Is there a nice way to do this using the Python library, or do I have to write it myself?


回答 0

您可以使用负数舍入整数:

>>> round(1234, -3)
1000.0

因此,如果您只需要最高有效数字:

>>> from math import log10, floor
>>> def round_to_1(x):
...   return round(x, -int(floor(log10(abs(x)))))
... 
>>> round_to_1(0.0232)
0.02
>>> round_to_1(1234243)
1000000.0
>>> round_to_1(13)
10.0
>>> round_to_1(4)
4.0
>>> round_to_1(19)
20.0

如果大于1,则可能需要将float转换为整数。

You can use negative numbers to round integers:

>>> round(1234, -3)
1000.0

Thus if you need only most significant digit:

>>> from math import log10, floor
>>> def round_to_1(x):
...   return round(x, -int(floor(log10(abs(x)))))
... 
>>> round_to_1(0.0232)
0.02
>>> round_to_1(1234243)
1000000.0
>>> round_to_1(13)
10.0
>>> round_to_1(4)
4.0
>>> round_to_1(19)
20.0

You’ll probably have to take care of turning float to integer if it’s bigger than 1.


回答 1

字符串格式的%g将格式化浮点数,并四舍五入到有效数字。有时会使用科学的“ e”表示法,因此将舍入的字符串转换回浮点数,然后通过%s字符串格式进行格式化。

>>> '%s' % float('%.1g' % 1234)
'1000'
>>> '%s' % float('%.1g' % 0.12)
'0.1'
>>> '%s' % float('%.1g' % 0.012)
'0.01'
>>> '%s' % float('%.1g' % 0.062)
'0.06'
>>> '%s' % float('%.1g' % 6253)
'6000.0'
>>> '%s' % float('%.1g' % 1999)
'2000.0'

%g in string formatting will format a float rounded to some number of significant figures. It will sometimes use ‘e’ scientific notation, so convert the rounded string back to a float then through %s string formatting.

>>> '%s' % float('%.1g' % 1234)
'1000'
>>> '%s' % float('%.1g' % 0.12)
'0.1'
>>> '%s' % float('%.1g' % 0.012)
'0.01'
>>> '%s' % float('%.1g' % 0.062)
'0.06'
>>> '%s' % float('%.1g' % 6253)
'6000.0'
>>> '%s' % float('%.1g' % 1999)
'2000.0'

回答 2

如果要使用除1个有效小数之外的其他数字(否则与Evgeny相同):

>>> from math import log10, floor
>>> def round_sig(x, sig=2):
...   return round(x, sig-int(floor(log10(abs(x))))-1)
... 
>>> round_sig(0.0232)
0.023
>>> round_sig(0.0232, 1)
0.02
>>> round_sig(1234243, 3)
1230000.0

If you want to have other than 1 significant decimal (otherwise the same as Evgeny):

>>> from math import log10, floor
>>> def round_sig(x, sig=2):
...   return round(x, sig-int(floor(log10(abs(x))))-1)
... 
>>> round_sig(0.0232)
0.023
>>> round_sig(0.0232, 1)
0.02
>>> round_sig(1234243, 3)
1230000.0

回答 3

f'{float(f"{i:.1g}"):g}'
# Or with Python <3.6,
'{:g}'.format(float('{:.1g}'.format(i)))

该解决方案与所有其他解决方案不同,因为:

  1. 恰好解决了OP问题
  2. 它并没有需要任何额外的包
  3. 需要任何用户定义的辅助功能数学运算

对于任意数量n的有效数字,可以使用:

print('{:g}'.format(float('{:.{p}g}'.format(i, p=n))))

测试:

a = [1234, 0.12, 0.012, 0.062, 6253, 1999, -3.14, 0., -48.01, 0.75]
b = ['{:g}'.format(float('{:.1g}'.format(i))) for i in a]
# b == ['1000', '0.1', '0.01', '0.06', '6000', '2000', '-3', '0', '-50', '0.8']

注意:使用此解决方案,不可能从输入动态地调整有效数字的数量,因为没有标准的方法来区分具有不同尾随零的数字(3.14 == 3.1400)。如果需要这样做,则需要非标准功能,例如精确度软件包中提供的功能。

f'{float(f"{i:.1g}"):g}'
# Or with Python <3.6,
'{:g}'.format(float('{:.1g}'.format(i)))

This solution is different from all of the others because:

  1. it exactly solves the OP question
  2. it does not need any extra package
  3. it does not need any user-defined auxiliary function or mathematical operation

For an arbitrary number n of significant figures, you can use:

print('{:g}'.format(float('{:.{p}g}'.format(i, p=n))))

Test:

a = [1234, 0.12, 0.012, 0.062, 6253, 1999, -3.14, 0., -48.01, 0.75]
b = ['{:g}'.format(float('{:.1g}'.format(i))) for i in a]
# b == ['1000', '0.1', '0.01', '0.06', '6000', '2000', '-3', '0', '-50', '0.8']

Note: with this solution, it is not possible to adapt the number of significant figures dynamically from the input because there is no standard way to distinguish numbers with different numbers of trailing zeros (3.14 == 3.1400). If you need to do so, then non-standard functions like the ones provided in the to-precision package are needed.


回答 4

我已经创建了可以满足您需求的高精度软件包。它使您可以为数字赋予或多或少的有效数字。

它还输出带有指定数量有效数字的标准,科学和工程符号。

在接受的答案中有一行

>>> round_to_1(1234243)
1000000.0

实际上指定了8个无花果。对于数字1234243,我的图书馆仅显示一个有效数字:

>>> from to_precision import to_precision
>>> to_precision(1234243, 1, 'std')
'1000000'
>>> to_precision(1234243, 1, 'sci')
'1e6'
>>> to_precision(1234243, 1, 'eng')
'1e6'

还将舍入最后一个有效数字,如果未指定符号,则可以自动选择要使用的符号:

>>> to_precision(599, 2)
'600'
>>> to_precision(1164, 2)
'1.2e3'

I have created the package to-precision that does what you want. It allows you to give your numbers more or less significant figures.

It also outputs standard, scientific, and engineering notation with a specified number of significant figures.

In the accepted answer there is the line

>>> round_to_1(1234243)
1000000.0

That actually specifies 8 sig figs. For the number 1234243 my library only displays one significant figure:

>>> from to_precision import to_precision
>>> to_precision(1234243, 1, 'std')
'1000000'
>>> to_precision(1234243, 1, 'sci')
'1e6'
>>> to_precision(1234243, 1, 'eng')
'1e6'

It will also round the last significant figure and can automatically choose what notation to use if a notation isn’t specified:

>>> to_precision(599, 2)
'600'
>>> to_precision(1164, 2)
'1.2e3'

回答 5

将整数四舍五入到1位有效数字的基本思想是将其转换为浮点数,该浮点数在该点之前1位数并四舍五入,然后将其转换回其原始整数大小。

为此,我们需要知道小于整数10的最大幂。为此,我们可以使用log 10功能的下限。

from math import log10, floor
def round_int(i,places):
    if i == 0:
        return 0
    isign = i/abs(i)
    i = abs(i)
    if i < 1:
        return 0
    max10exp = floor(log10(i))
    if max10exp+1 < places:
        return i
    sig10pow = 10**(max10exp-places+1)
    floated = i*1.0/sig10pow
    defloated = round(floated)*sig10pow
    return int(defloated*isign)

To round an integer to 1 significant figure the basic idea is to convert it to a floating point with 1 digit before the point and round that, then convert it back to its original integer size.

To do this we need to know the largest power of 10 less than the integer. We can use floor of the log 10 function for this.

from math import log10, floor
def round_int(i,places):
    if i == 0:
        return 0
    isign = i/abs(i)
    i = abs(i)
    if i < 1:
        return 0
    max10exp = floor(log10(i))
    if max10exp+1 < places:
        return i
    sig10pow = 10**(max10exp-places+1)
    floated = i*1.0/sig10pow
    defloated = round(floated)*sig10pow
    return int(defloated*isign)

回答 6

为了直接回答这个问题,这是我使用R函数命名的版本:

import math

def signif(x, digits=6):
    if x == 0 or not math.isfinite(x):
        return x
    digits -= math.ceil(math.log10(abs(x)))
    return round(x, digits)

我发布此答案的主要原因是评论抱怨“ 0.075”舍入为0.07而不是0.08。如“新手C”所指出的,这是由于具有有限精度和以2为底的表示形式的浮点运算的组合。实际上可以表示的最接近0.075的数字要小一些,因此舍入的结果与您可能天真地期望的不同。

还要注意,这适用于任何非十进制浮点算法的使用,例如C和Java都有相同的问题。

为了更详细地显示,我们要求Python将数字格式化为“十六进制”格式:

0.075.hex()

这给了我们:0x1.3333333333333p-4。这样做的原因是正常的十进制表示形式通常涉及舍入,因此不是计算机实际“看到”数字的方式。如果您不习惯这种格式,那么Python docsC standard是两个有用的参考。

为了说明这些数字是如何工作的,我们可以通过以下操作回到起点:

0x13333333333333 / 16**13 * 2**-4

应该打印出来0.07516**13是因为小数点后有13个十六进制数字,并且2**-4是因为十六进制指数是以2为底的。

现在我们有了关于浮点数表示方式的一些想法,可以使用decimal模块为我们提供更多的精度,向我们展示发生了什么事情:

from decimal import Decimal

Decimal(0x13333333333333) / 16**13 / 2**4

给:0.07499999999999999722444243844并希望解释为什么round(0.075, 2)评估0.07

To directly answer the question, here’s my version using naming from the R function:

import math

def signif(x, digits=6):
    if x == 0 or not math.isfinite(x):
        return x
    digits -= math.ceil(math.log10(abs(x)))
    return round(x, digits)

My main reason for posting this answer are the comments complaining that “0.075” rounds to 0.07 rather than 0.08. This is due, as pointed out by “Novice C”, to a combination of floating point arithmetic having both finite precision and a base-2 representation. The nearest number to 0.075 that can actually be represented is slightly smaller, hence rounding comes out differently than you might naively expect.

Also note that this applies to any use of non-decimal floating point arithmetic, e.g. C and Java both have the same issue.

To show in more detail, we ask Python to format the number in “hex” format:

0.075.hex()

which gives us: 0x1.3333333333333p-4. The reason for doing this is that the normal decimal representation often involves rounding and hence is not how the computer actually “sees” the number. If you’re not used to this format, a couple of useful references are the Python docs and the C standard.

To show how these numbers work a bit, we can get back to our starting point by doing:

0x13333333333333 / 16**13 * 2**-4

which should should print out 0.075. 16**13 is because there are 13 hexadecimal digits after the decimal point, and 2**-4 is because hex exponents are base-2.

Now we have some idea of how floats are represented we can use the decimal module to give us some more precision, showing us what’s going on:

from decimal import Decimal

Decimal(0x13333333333333) / 16**13 / 2**4

giving: 0.07499999999999999722444243844 and hopefully explaining why round(0.075, 2) evaluates to 0.07


回答 7

def round_to_n(x, n):
    if not x: return 0
    power = -int(math.floor(math.log10(abs(x)))) + (n - 1)
    factor = (10 ** power)
    return round(x * factor) / factor

round_to_n(0.075, 1)      # 0.08
round_to_n(0, 1)          # 0
round_to_n(-1e15 - 1, 16) # 1000000000000001.0

希望能充分利用以上所有答案中的最佳答案(减去能够将其表示为一行lambda的意思;)。尚未探索,请随时编辑此答案:

round_to_n(1e15 + 1, 11)  # 999999999999999.9
def round_to_n(x, n):
    if not x: return 0
    power = -int(math.floor(math.log10(abs(x)))) + (n - 1)
    factor = (10 ** power)
    return round(x * factor) / factor

round_to_n(0.075, 1)      # 0.08
round_to_n(0, 1)          # 0
round_to_n(-1e15 - 1, 16) # 1000000000000001.0

Hopefully taking the best of all the answers above (minus being able to put it as a one line lambda ;) ). Haven’t explored yet, feel free to edit this answer:

round_to_n(1e15 + 1, 11)  # 999999999999999.9

回答 8

我修改了indgar的解决方案,以处理负数和小数(包括零)。

from math import log10, floor
def round_sig(x, sig=6, small_value=1.0e-9):
    return round(x, sig - int(floor(log10(max(abs(x), abs(small_value))))) - 1)

I modified indgar’s solution to handle negative numbers and small numbers (including zero).

from math import log10, floor
def round_sig(x, sig=6, small_value=1.0e-9):
    return round(x, sig - int(floor(log10(max(abs(x), abs(small_value))))) - 1)

回答 9

如果您想不涉及字符串而四舍五入,我发现上面的链接中隐藏了该链接:

http://code.activestate.com/lists/python-tutor/70739/

给我最好的印象 然后,当您使用任何字符串格式的描述符进行打印时,您将获得一个合理的输出,并且可以将数字表示形式用于其他计算目的。

链接上的代码由三部分组成:def,doc和return。它有一个错误:您需要检查对数是否爆炸。那很容易。将输入与进行比较sys.float_info.min。完整的解决方案是:

import sys,math

def tidy(x, n):
"""Return 'x' rounded to 'n' significant digits."""
y=abs(x)
if y <= sys.float_info.min: return 0.0
return round( x, int( n-math.ceil(math.log10(y)) ) )

它适用于任何标量数值,float如果由于某种原因需要移动响应,n可以为a 。您实际上可以将限制提高到:

sys.float_info.min*sys.float_info.epsilon

如果您出于某些原因正在使用极小的值,则不会引发错误。

If you want to round without involving strings, the link I found buried in the comments above:

http://code.activestate.com/lists/python-tutor/70739/

strikes me as best. Then when you print with any string formatting descriptors, you get a reasonable output, and you can use the numeric representation for other calculation purposes.

The code at the link is a three liner: def, doc, and return. It has a bug: you need to check for exploding logarithms. That is easy. Compare the input to sys.float_info.min. The complete solution is:

import sys,math

def tidy(x, n):
"""Return 'x' rounded to 'n' significant digits."""
y=abs(x)
if y <= sys.float_info.min: return 0.0
return round( x, int( n-math.ceil(math.log10(y)) ) )

It works for any scalar numeric value, and n can be a float if you need to shift the response for some reason. You can actually push the limit to:

sys.float_info.min*sys.float_info.epsilon

without provoking an error, if for some reason you are working with miniscule values.


回答 10

我想不出任何能够立即解决的问题。但是对于浮点数来说,它处理得很好。

>>> round(1.2322, 2)
1.23

整数比较棘手。它们不会以10为基数存储在内存中,因此重要的地方并不是自然而然的事情。但是,一旦成为字符串,实现起来就很简单了。

或对于整数:

>>> def intround(n, sigfigs):
...   n = str(n)
...   return n[:sigfigs] + ('0' * (len(n)-(sigfigs)))

>>> intround(1234, 1)
'1000'
>>> intround(1234, 2)

如果您想创建一个可以处理任何数字的函数,我的偏好是将它们都转换为字符串,并寻找一个小数位来决定要做什么:

>>> def roundall1(n, sigfigs):
...   n = str(n)
...   try:
...     sigfigs = n.index('.')
...   except ValueError:
...     pass
...   return intround(n, sigfigs)

另一种选择是检查类型。这将不太灵活,并且可能无法与其他数字(例如Decimal对象)很好地配合使用:

>>> def roundall2(n, sigfigs):
...   if type(n) is int: return intround(n, sigfigs)
...   else: return round(n, sigfigs)

I can’t think of anything that would be able to handle this out of the box. But it’s fairly well handled for floating point numbers.

>>> round(1.2322, 2)
1.23

Integers are trickier. They’re not stored as base 10 in memory, so significant places isn’t a natural thing to do. It’s fairly trivial to implement once they’re a string though.

Or for integers:

>>> def intround(n, sigfigs):
...   n = str(n)
...   return n[:sigfigs] + ('0' * (len(n)-(sigfigs)))

>>> intround(1234, 1)
'1000'
>>> intround(1234, 2)

If you would like to create a function that handles any number, my preference would be to convert them both to strings and look for a decimal place to decide what to do:

>>> def roundall1(n, sigfigs):
...   n = str(n)
...   try:
...     sigfigs = n.index('.')
...   except ValueError:
...     pass
...   return intround(n, sigfigs)

Another option is to check for type. This will be far less flexible, and will probably not play nicely with other numbers such as Decimal objects:

>>> def roundall2(n, sigfigs):
...   if type(n) is int: return intround(n, sigfigs)
...   else: return round(n, sigfigs)

回答 11

给出的答案是给出的最好的答案,但是它有很多限制,并且在技术上没有正确的有效数字。

numpy.format_float_positional直接支持所需的行为。以下片段将浮点数x格式化为4个有效数字,并取消了科学计数法。

import numpy as np
x=12345.6
np.format_float_positional(x, precision=4, unique=False, fractional=False, trim='k')
> 12340.

The posted answer was the best available when given, but it has a number of limitations and does not produce technically correct significant figures.

numpy.format_float_positional supports the desired behaviour directly. The following fragment returns the float x formatted to 4 significant figures, with scientific notation suppressed.

import numpy as np
x=12345.6
np.format_float_positional(x, precision=4, unique=False, fractional=False, trim='k')
> 12340.

回答 12

我也遇到了这个问题,但是我需要控制舍入类型。因此,我编写了一个快速函数(请参见下面的代码),该函数可以将值,舍入类型和所需的有效数字考虑在内。

import decimal
from math import log10, floor

def myrounding(value , roundstyle='ROUND_HALF_UP',sig = 3):
    roundstyles = [ 'ROUND_05UP','ROUND_DOWN','ROUND_HALF_DOWN','ROUND_HALF_UP','ROUND_CEILING','ROUND_FLOOR','ROUND_HALF_EVEN','ROUND_UP']

    power =  -1 * floor(log10(abs(value)))
    value = '{0:f}'.format(value) #format value to string to prevent float conversion issues
    divided = Decimal(value) * (Decimal('10.0')**power) 
    roundto = Decimal('10.0')**(-sig+1)
    if roundstyle not in roundstyles:
        print('roundstyle must be in list:', roundstyles) ## Could thrown an exception here if you want.
    return_val = decimal.Decimal(divided).quantize(roundto,rounding=roundstyle)*(decimal.Decimal(10.0)**-power)
    nozero = ('{0:f}'.format(return_val)).rstrip('0').rstrip('.') # strips out trailing 0 and .
    return decimal.Decimal(nozero)


for x in list(map(float, '-1.234 1.2345 0.03 -90.25 90.34543 9123.3 111'.split())):
    print (x, 'rounded UP: ',myrounding(x,'ROUND_UP',3))
    print (x, 'rounded normal: ',myrounding(x,sig=3))

I ran into this as well but I needed control over the rounding type. Thus, I wrote a quick function (see code below) that can take value, rounding type, and desired significant digits into account.

import decimal
from math import log10, floor

def myrounding(value , roundstyle='ROUND_HALF_UP',sig = 3):
    roundstyles = [ 'ROUND_05UP','ROUND_DOWN','ROUND_HALF_DOWN','ROUND_HALF_UP','ROUND_CEILING','ROUND_FLOOR','ROUND_HALF_EVEN','ROUND_UP']

    power =  -1 * floor(log10(abs(value)))
    value = '{0:f}'.format(value) #format value to string to prevent float conversion issues
    divided = Decimal(value) * (Decimal('10.0')**power) 
    roundto = Decimal('10.0')**(-sig+1)
    if roundstyle not in roundstyles:
        print('roundstyle must be in list:', roundstyles) ## Could thrown an exception here if you want.
    return_val = decimal.Decimal(divided).quantize(roundto,rounding=roundstyle)*(decimal.Decimal(10.0)**-power)
    nozero = ('{0:f}'.format(return_val)).rstrip('0').rstrip('.') # strips out trailing 0 and .
    return decimal.Decimal(nozero)


for x in list(map(float, '-1.234 1.2345 0.03 -90.25 90.34543 9123.3 111'.split())):
    print (x, 'rounded UP: ',myrounding(x,'ROUND_UP',3))
    print (x, 'rounded normal: ',myrounding(x,sig=3))

回答 13

使用python 2.6+ 新样式格式(不建议使用%-style):

>>> "{0}".format(float("{0:.1g}".format(1216)))
'1000.0'
>>> "{0}".format(float("{0:.1g}".format(0.00356)))
'0.004'

在python 2.7+中,您可以省略前导0s。

Using python 2.6+ new-style formatting (as %-style is deprecated):

>>> "{0}".format(float("{0:.1g}".format(1216)))
'1000.0'
>>> "{0}".format(float("{0:.1g}".format(0.00356)))
'0.004'

In python 2.7+ you can omit the leading 0s.


回答 14

如果数字大于10 **(-decimal_positions),则此函数进行正常的回合,否则增加更多的小数,直到达到有意义的小数位数为止:

def smart_round(x, decimal_positions):
    dp = - int(math.log10(abs(x))) if x != 0.0 else int(0)
    return round(float(x), decimal_positions + dp if dp > 0 else decimal_positions)

希望能帮助到你。

This function does a normal round if the number is bigger than 10**(-decimal_positions), otherwise adds more decimal until the number of meaningful decimal positions is reached:

def smart_round(x, decimal_positions):
    dp = - int(math.log10(abs(x))) if x != 0.0 else int(0)
    return round(float(x), decimal_positions + dp if dp > 0 else decimal_positions)

Hope it helps.


回答 15

https://stackoverflow.com/users/1391441/gabriel,以下内容是否解决了您对rnd(.075,1)的担忧?警告:以浮点数形式返回值

def round_to_n(x, n):
    fmt = '{:1.' + str(n) + 'e}'    # gives 1.n figures
    p = fmt.format(x).split('e')    # get mantissa and exponent
                                    # round "extra" figure off mantissa
    p[0] = str(round(float(p[0]) * 10**(n-1)) / 10**(n-1))
    return float(p[0] + 'e' + p[1]) # convert str to float

>>> round_to_n(750, 2)
750.0
>>> round_to_n(750, 1)
800.0
>>> round_to_n(.0750, 2)
0.075
>>> round_to_n(.0750, 1)
0.08
>>> math.pi
3.141592653589793
>>> round_to_n(math.pi, 7)
3.141593

https://stackoverflow.com/users/1391441/gabriel, does the following address your concern about rnd(.075, 1)? Caveat: returns value as a float

def round_to_n(x, n):
    fmt = '{:1.' + str(n) + 'e}'    # gives 1.n figures
    p = fmt.format(x).split('e')    # get mantissa and exponent
                                    # round "extra" figure off mantissa
    p[0] = str(round(float(p[0]) * 10**(n-1)) / 10**(n-1))
    return float(p[0] + 'e' + p[1]) # convert str to float

>>> round_to_n(750, 2)
750.0
>>> round_to_n(750, 1)
800.0
>>> round_to_n(.0750, 2)
0.075
>>> round_to_n(.0750, 1)
0.08
>>> math.pi
3.141592653589793
>>> round_to_n(math.pi, 7)
3.141593

回答 16

这将返回一个字符串,以便结果不包含小数部分,并且正确显示了E表示法中否则会出现的较小值:

def sigfig(x, num_sigfig):
    num_decplace = num_sigfig - int(math.floor(math.log10(abs(x)))) - 1
    return '%.*f' % (num_decplace, round(x, num_decplace))

This returns a string, so that results without fractional parts, and small values which would otherwise appear in E notation are shown correctly:

def sigfig(x, num_sigfig):
    num_decplace = num_sigfig - int(math.floor(math.log10(abs(x)))) - 1
    return '%.*f' % (num_decplace, round(x, num_decplace))

回答 17

给定一个如此彻底回答的问题,为什么不添加另一个

尽管上面的许多内容是可比的,但这更适合我的审美观

import numpy as np

number=-456.789
significantFigures=4

roundingFactor=significantFigures - int(np.floor(np.log10(np.abs(number)))) - 1
rounded=np.round(number, roundingFactor)

string=rounded.astype(str)

print(string)

这适用于单个数字和numpy数组,对于负数应该可以正常工作。

我们可能还要增加一个附加步骤-即使四舍五入为整数,np.round()也会返回一个十进制数(即,对于ificantFigures = 2,我们可能期望返回-460,但相反会得到-460.0)。我们可以添加此步骤以更正此问题:

if roundingFactor<=0:
    rounded=rounded.astype(int)

不幸的是,最后一步不适用于数字数组-亲爱的读者,我会把这个留给您看看是否需要。

Given a question so thoroughly answered why not add another

This suits my aesthetic a little better, though many of the above are comparable

import numpy as np

number=-456.789
significantFigures=4

roundingFactor=significantFigures - int(np.floor(np.log10(np.abs(number)))) - 1
rounded=np.round(number, roundingFactor)

string=rounded.astype(str)

print(string)

This works for individual numbers and numpy arrays, and should function fine for negative numbers.

There’s one additional step we might add – np.round() returns a decimal number even if rounded is an integer (i.e. for significantFigures=2 we might expect to get back -460 but instead we get -460.0). We can add this step to correct for that:

if roundingFactor<=0:
    rounded=rounded.astype(int)

Unfortunately, this final step won’t work for an array of numbers – I’ll leave that to you dear reader to figure out if you need.


回答 18

sigfig包/库盖这一点。后安装,你可以做到以下几点:

>>> from sigfig import round
>>> round(1234, 1)
1000
>>> round(0.12, 1)
0.1
>>> round(0.012, 1)
0.01
>>> round(0.062, 1)
0.06
>>> round(6253, 1)
6000
>>> round(1999, 1)
2000

The sigfig package/library covers this. After installing you can do the following:

>>> from sigfig import round
>>> round(1234, 1)
1000
>>> round(0.12, 1)
0.1
>>> round(0.012, 1)
0.01
>>> round(0.062, 1)
0.06
>>> round(6253, 1)
6000
>>> round(1999, 1)
2000

回答 19

import math

  def sig_dig(x, n_sig_dig):
      num_of_digits = len(str(x).replace(".", ""))
      if n_sig_dig >= num_of_digits:
          return x
      n = math.floor(math.log10(x) + 1 - n_sig_dig)
      result = round(10 ** -n * x) * 10 ** n
      return float(str(result)[: n_sig_dig + 1])


    >>> sig_dig(1234243, 3)
    >>> sig_dig(243.3576, 5)

        1230.0
        243.36
import math

  def sig_dig(x, n_sig_dig):
      num_of_digits = len(str(x).replace(".", ""))
      if n_sig_dig >= num_of_digits:
          return x
      n = math.floor(math.log10(x) + 1 - n_sig_dig)
      result = round(10 ** -n * x) * 10 ** n
      return float(str(result)[: n_sig_dig + 1])


    >>> sig_dig(1234243, 3)
    >>> sig_dig(243.3576, 5)

        1230.0
        243.36

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