blocks|key|1602024|text|'%25f'+%25+(x/y)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1602025|但你需要自己管理精确度。例如，|unstyled|1602026|'%25f'+%25+(1/10**8)|1602027|将仅显示零。|1602028|details+are+in+the+docs|offset|length|1602029|或者用于Python3+the+equivalent+old+formatting或newer+style+formatting|1602030|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/string.html#formatstrings|1|http://docs.python.org/py3k/library/stdtypes.html#old-string-formatting-operations|2|http://docs.python.org/py3k/library/string.html#string-formatting^0|0|0|0|0|0|N|0|0|C|T|1|16|M|2|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|13|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|14|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|15|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|16|8|@]|9|@$M|17|N|18|1|19]]|A|$]]|$1|O|3|P|5|F|7|1A|8|@]|9|@$M|1B|N|1C|1|1D]|$M|1E|N|1F|1|1G]]|A|$]]|$1|Q|3|-4|5|F|7|1H|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]|Y|$5|T|U|V|A|$W|Z]]|10|$5|T|U|V|A|$W|11]]]]

<pre><code>'%f' % (x/y)
</code></pre>
but you need to manage precision yourself. e.g.,
<pre><code>'%f' % (1/10**8)
</code></pre>
will display zeros only. 
<a href="https://docs.python.org/3/library/string.html#formatstrings" rel="nofollow noreferrer">details are in the docs</a>
Or for Python 3 <a href="http://docs.python.org/py3k/library/stdtypes.html#old-string-formatting-operations" rel="nofollow noreferrer">the equivalent old formatting</a> or the <a href="http://docs.python.org/py3k/library/string.html#string-formatting" rel="nofollow noreferrer">newer style formatting</a>

blocks|key|1602107|text|除了SG的答案之外，您还可以使用Decimal模块：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1602108|from+decimal+import+Decimal
x+=+str(Decimal(1)+/+Decimal(10000))

#+x+is+a+string+'0.0001'|code-block|syntax|javascript|1602109|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

In addition to SG's answer, you can also use the Decimal module:

<pre><code>from decimal import Decimal
x = str(Decimal(1) / Decimal(10000))

# x is a string '0.0001'
</code></pre>

blocks|key|1602138|text|使用较新版本的Python2.6或更高版本，您可以使用''.format()来完成@SilentGhost建议的操作：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1602139|'{0:f}'.format(x/y)|code-block|syntax|javascript|1602140|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/string.html#format-string-syntax^0|R|B|R|B|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

With newer versions of Python (2.6 and later), you can use <a href="http://docs.python.org/library/string.html#format-string-syntax"><code>''.format()</code></a> to accomplish what @SilentGhost suggested:

<pre><code>'{0:f}'.format(x/y)
</code></pre>

blocks|key|1602188|text|如果是string，则在其上使用内置的float来执行转换，例如：print(+"%25.5f"+%25+float("1.43572e-03"))答案：0.00143572|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1602189|entityMap^0|3|6|J|5|X|11|21|A|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|G|$]]

If it is a <code>string</code> then use the built in <code>float</code> on it to do the conversion for instance:
<code>print( "%.5f" % float("1.43572e-03"))</code>
answer:<code>0.00143572</code>

blocks|key|1602231|text|使用较新版本的''.format+(还要记住指定您希望显示的.之后的位数，这取决于浮点数的小小)。请参阅此示例：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1602232|>>>+a+=+-7.1855143557448603e-17
>>>+'{:f}'.format(a)
'-0.000000'|code-block|syntax|javascript|1602233|如上所示，默认为6位！这对我们的案例示例没有帮助，所以我们可以使用下面这样的内容：|1602234|>>>+'{:.20f}'.format(a)
'-0.00000000000000007186'|1602235|更新|1602236|从Python3.6开始，可以使用新的formatted+string+literal简化这一过程，如下所示：|1602237|>>>+f'{a:.20f}'
'-0.00000000000000007186'|1602238|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/whatsnew/3.6.html#whatsnew36-pep498^0|7|9|U|1|0|0|0|0|0|J|O|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|18|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|19|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1A|8|@]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1B|8|@]|D|@$9|1C|A|1D|1|1E]]|E|$]]|$1|S|3|T|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|U|3|-4|5|6|7|1G|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11]]]]

Using the newer version <code>''.format</code> (also remember to specify how many digit after the <code>.</code> you wish to display, this depends on how small is the floating number). See this example:

<pre><code>&gt;&gt;&gt; a = -7.1855143557448603e-17
&gt;&gt;&gt; '{:f}'.format(a)
'-0.000000'
</code></pre>

as shown above, default is 6 digits! This is not helpful for our case example, so instead we could use something like this:

<pre><code>&gt;&gt;&gt; '{:.20f}'.format(a)
'-0.00000000000000007186'
</code></pre>

<h2>Update</h2>

Starting in Python 3.6, this can be simplified with the new <a href="https://docs.python.org/3/whatsnew/3.6.html#whatsnew36-pep498" rel="noreferrer">formatted string literal</a>, as follows:

<pre><code>&gt;&gt;&gt; f'{a:.20f}'
'-0.00000000000000007186'
</code></pre>

blocks|key|1387416|text|这对任何指数都有效：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1387417|def+getExpandedScientificNotation(flt):
++++str_vals+=+str(flt).split('e')
++++coef+=+float(str_vals[0])
++++exp+=+int(str_vals[1])
++++return_val+=+''
++++if+int(exp)+>+0:
++++++++return_val+%2B=+str(coef).replace('.',+'')
++++++++return_val+%2B=+''.join(['0'+for+_+in+range(0,+abs(exp+-+len(str(coef).split('.')[1])))])
++++elif+int(exp)+<+0:
++++++++return_val+%2B=+'0.'
++++++++return_val+%2B=+''.join(['0'+for+_+in+range(0,+abs(exp)+-+1)])
++++++++return_val+%2B=+str(coef).replace('.',+'')
++++return+return_val|code-block|syntax|javascript|1387418|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This will work for any exponent:

<pre><code>def getExpandedScientificNotation(flt):
 str_vals = str(flt).split('e')
 coef = float(str_vals[0])
 exp = int(str_vals[1])
 return_val = ''
 if int(exp) &gt; 0:
 return_val += str(coef).replace('.', '')
 return_val += ''.join(['0' for _ in range(0, abs(exp - len(str(coef).split('.')[1])))])
 elif int(exp) &lt; 0:
 return_val += '0.'
 return_val += ''.join(['0' for _ in range(0, abs(exp) - 1)])
 return_val += str(coef).replace('.', '')
 return return_val
</code></pre>

blocks|key|1387470|text|这是使用Captain+Cucumber's+answer，但增加了2个。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1387471|1)允许函数获取非科学记数法数字，并按原样返回它们(因此，您可以抛出大量输入，其中一些数字是0.00003123+vs+3.123e-05，并且函数仍然有效。|1387472|2)增加对负数的支持。(在原始函数中，负数的结尾为-1.08904e-05中的0.0000-108904+)|1387473|def+getExpandedScientificNotation(flt):
++++was_neg+=+False
++++if+not+("e"+in+flt):
++++++++return+flt
++++if+flt.startswith('-'):
++++++++flt+=+flt[1:]
++++++++was_neg+=+True+
++++str_vals+=+str(flt).split('e')
++++coef+=+float(str_vals[0])
++++exp+=+int(str_vals[1])
++++return_val+=+''
++++if+int(exp)+>+0:
++++++++return_val+%2B=+str(coef).replace('.',+'')
++++++++return_val+%2B=+''.join(['0'+for+_+in+range(0,+abs(exp+-+len(str(coef).split('.')[1])))])
++++elif+int(exp)+<+0:
++++++++return_val+%2B=+'0.'
++++++++return_val+%2B=+''.join(['0'+for+_+in+range(0,+abs(exp)+-+1)])
++++++++return_val+%2B=+str(coef).replace('.',+'')
++++if+was_neg:
++++++++return_val='-'%2Breturn_val
++++return+return_val|code-block|syntax|javascript|1387474|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/45604186/8282282^0|4|P|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|6|7|Y|8|@]|9|@]|C|$]]|$1|F|3|G|5|6|7|Z|8|@]|9|@]|C|$]]|$1|H|3|I|5|J|7|10|8|@]|9|@]|C|$K|L]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

This is using <a href="https://stackoverflow.com/a/45604186/8282282">Captain Cucumber's answer</a>, but with 2 additions.

1) allowing the function to get non scientific notation numbers and just return them as is (so you can throw a lot of input that some of the numbers are 0.00003123 vs 3.123e-05 and still have function work.

2) added support for negative numbers. (in original function, a negative number would end up like 0.0000-108904 from -1.08904e-05)

<pre><code>def getExpandedScientificNotation(flt):
 was_neg = False
 if not ("e" in flt):
 return flt
 if flt.startswith('-'):
 flt = flt[1:]
 was_neg = True 
 str_vals = str(flt).split('e')
 coef = float(str_vals[0])
 exp = int(str_vals[1])
 return_val = ''
 if int(exp) &gt; 0:
 return_val += str(coef).replace('.', '')
 return_val += ''.join(['0' for _ in range(0, abs(exp - len(str(coef).split('.')[1])))])
 elif int(exp) &lt; 0:
 return_val += '0.'
 return_val += ''.join(['0' for _ in range(0, abs(exp) - 1)])
 return_val += str(coef).replace('.', '')
 if was_neg:
 return_val='-'+return_val
 return return_val
</code></pre>

blocks|key|1387502|text|在使用3.6.4时，我遇到了一个类似的问题，在使用以下代码时，输出文件中的数字会被随机设置为科学记数法的格式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1387503|fout.write('someFloats:+{0:0.8},{1:0.8},{2:0.8}'.format(someFloat[0],+someFloat[1],+someFloat[2]))|code-block|syntax|javascript|1387504|我所要做的就是添加'f'：|1387505|fout.write('someFloats:+{0:0.8f},{1:0.8f},{2:0.8f}'.format(someFloat[0],+someFloat[1],+someFloat[2]))|1387506|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Using 3.6.4, I was having a similar problem that randomly, a number in the output file would be formatted with scientific notation when using this:

<pre><code>fout.write('someFloats: {0:0.8},{1:0.8},{2:0.8}'.format(someFloat[0], someFloat[1], someFloat[2]))
</code></pre>

All that I had to do to fix it was to add 'f':

<pre><code>fout.write('someFloats: {0:0.8f},{1:0.8f},{2:0.8f}'.format(someFloat[0], someFloat[1], someFloat[2]))
</code></pre>

blocks|key|1602379|text|另一种选择是，如果您使用的是pandas，并且想要取消所有浮点数的科学记数法，则可以调整pandas选项。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1602380|import+pandas+as+pd
pd.options.display.float_format+=+'{:.2f}'.format|code-block|syntax|javascript|1602381|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Another option, if you are using pandas and would like to suppress scientific notation for all floats, is to adjust the pandas options.

<pre><code>import pandas as pd
pd.options.display.float_format = '{:.2f}'.format
</code></pre>

blocks|key|1387558|text|在3.6版本中(可能也适用于更老的3.x版本)，这是我的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1387559|import+locale
locale.setlocale(locale.LC_ALL,+'')

def+number_format(n,+dec_precision=4):
++++precision+=+len(str(round(n)))+%2B+dec_precision
++++return+format(float(n),+f'.{precision}n')|code-block|syntax|javascript|1387560|precision计算的目的是确保我们有足够的精度来避免使用科学记数法(默认精度仍然是6)。|offset|length|style|CODE|1387561|dec_precision参数增加了用于小数点的额外精度。由于这使用了n格式，因此不会添加无关紧要的零(与f格式不同)。n还将负责呈现不带小数的已舍入的整数。|1387562|n确实需要float输入，因此需要强制转换。|1387563|entityMap^0|0|0|0|9|0|0|D|Z|1|1H|1|1O|1|0|0|1|5|5|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@$I|Y|J|Z|K|L]|$I|10|J|11|K|L]|$I|12|J|13|K|L]|$I|14|J|15|K|L]]|9|@]|A|$]]|$1|O|3|P|5|6|7|16|8|@$I|17|J|18|K|L]|$I|19|J|1A|K|L]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|R|$]]

As of 3.6 (probably works with slightly older 3.x as well), this is my solution:

<pre><code>import locale
locale.setlocale(locale.LC_ALL, '')

def number_format(n, dec_precision=4):
 precision = len(str(round(n))) + dec_precision
 return format(float(n), f'.{precision}n')
</code></pre>

The purpose of the <code>precision</code> calculation is to ensure we have enough precision to keep out of scientific notation (default precision is still 6).

The <code>dec_precision</code> argument adds additional precision to use for decimal points. Since this makes use of the <code>n</code> format, no insignificant zeros will be added (unlike <code>f</code> formats). <code>n</code> also will take care of rendering already-round integers without a decimal.

<code>n</code> does require <code>float</code> input, thus the cast.

blocks|key|1387604|text|上面的大多数答案都需要您指定精度。但是，如果你想像这样显示浮点数，没有不必要的零该怎么办：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1387605|1
0.1
0.01
0.001
0.0001
0.00001
0.000001
0.000000000001|code-block|syntax|javascript|1387606|numpy有一个答案：np.format_float_positional|offset|length|style|CODE|1387607|import+numpy+as+np

def+format_float(num):
++++return+np.format_float_positional(num,+trim='-')|1387608|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.scipy.org/doc/numpy/reference/generated/numpy.format_float_positional.html^0|0|0|0|5|B|Q|B|Q|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]|$I|11|J|12|K|L]]|9|@$I|13|J|14|1|15]]|A|$]]|$1|M|3|N|5|D|7|16|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

Most of the answers above require you to specify precision. But what if you want to display floats like this, with no unnecessary zeros: 

<pre><code>1
0.1
0.01
0.001
0.0001
0.00001
0.000001
0.000000000001
</code></pre>

<code>numpy</code> has an answer: <a href="https://docs.scipy.org/doc/numpy/reference/generated/numpy.format_float_positional.html" rel="noreferrer"><code>np.format_float_positional</code></a>

<pre><code>import numpy as np

def format_float(num):
 return np.format_float_positional(num, trim='-')
</code></pre>

blocks|key|1387663|text|一种更简单的解决方案，将浮点数显示为任意数量的有效数字。这里不需要numpy或列表理解：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1387664|def+sig(num,+digits=3):
++++"Return+number+formatted+for+significant+digits"
++++if+num+==+0:
++++++++return+0
++++negative+=+'-'+if+num+<+0+else+''
++++num+=+abs(float(num))
++++power+=+math.log(num,+10)
++++if+num+<+1:
++++++++step+=+int(10**(-int(power)+%2B+digits)+*+num)
++++++++return+negative+%2B+'0.'+%2B+'0'+*+-int(power)+%2B+str(int(step)).rstrip('0')
++++elif+power+<+digits+-+1:
++++++++return+negative+%2B+('{0:.'+%2B+str(digits)+%2B+'g}').format(num)
++++else:
++++++++return+negative+%2B+str(int(num))|code-block|syntax|javascript|1387665|我在示例中去掉了尾随的0并显示了完整的整数：sig(31415.9)+=+31415而不是31400。如果您对此不感兴趣，请随时修改代码。|1387666|测试：|1387667|for+power+in+range(-8,8):
++++num+=+math.pi+*+10**power
++++print(str(num).ljust(25),+sig(num))|1387668|entityMap^0|X|5|0|0|M|K|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|Z|8|@]|D|@]|E|$]]|$1|O|3|P|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|Q|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|R|$]]

A simpler solution to display a float to an arbitrary number of significant digits. No <code>numpy</code> or list comprehensions required here:
<pre><code>def sig(num, digits=3):
 &quot;Return number formatted for significant digits&quot;
 if num == 0:
 return 0
 negative = '-' if num &lt; 0 else ''
 num = abs(float(num))
 power = math.log(num, 10)
 if num &lt; 1:
 step = int(10**(-int(power) + digits) * num)
 return negative + '0.' + '0' * -int(power) + str(int(step)).rstrip('0')
 elif power &lt; digits - 1:
 return negative + ('{0:.' + str(digits) + 'g}').format(num)
 else:
 return negative + str(int(num))
</code></pre>
I'm stripping trailing 0s and displaying full integers in the example: <code>sig(31415.9) = 31415</code> instead of 31400. Feel free to modify the code if that's not something you're into.
Testing:
<pre><code>for power in range(-8,8):
 num = math.pi * 10**power
 print(str(num).ljust(25), sig(num))
</code></pre>

blocks|key|1602507|text|您可以使用内置的format函数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1602508|>>>+a+=+-3.42142141234123e-15
>>>+format(a,+'f')
'-0.000000'
>>>+format(a,+'.50f')+#+Or+you+can+specify+precision
'-0.00000000000000342142141234122994048466990874926279'|code-block|syntax|javascript|1602509|entityMap^0|8|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

You can use the built-in <code>format</code> function.
<pre><code>&gt;&gt;&gt; a = -3.42142141234123e-15
&gt;&gt;&gt; format(a, 'f')
'-0.000000'
&gt;&gt;&gt; format(a, '.50f') # Or you can specify precision
'-0.00000000000000342142141234122994048466990874926279'
</code></pre>

blocks|key|1602533|text|在numpy数组的情况下，您可以使用suppress命令抑制为|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1602534|import+numpy+as+np
np.set_printoptions(suppress=True)|code-block|syntax|javascript|1602535|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

In case of numpy arrays you can suppress with suppress command as
<pre><code>import numpy as np
np.set_printoptions(suppress=True)
</code></pre>

Here's my code:

<pre><code>x = 1.0
y = 100000.0 
print x/y
</code></pre>

My quotient displays as <code>1.00000e-05</code>.

Is there any way to suppress scientific notation and make it display as
<code>0.00001</code>? I'm going to use the result as a string.

How to suppress scientific notation when printing float values?

下面是我的代码：x = 1.0y = 100000.0    print x/y我的商显示为1.00000e-05。有没有办法抑制科学记数法并使其显示为0.00001？我将使用结果作为字符串。

问如何在打印浮点值时抑制科学记数法？
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