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You should install the python3 and python3-doc packages and read the <a href="file:///usr/share/doc/python3-doc/html/tutorial/floatingpoint.html"> Python documentation for the Python3 version packaged in this release</a>. </div> <div class="related" role="navigation" aria-label="related navigation"> <h3>Navigation</h3> <ul> <li class="right" style="margin-right: 10px"> <a href="../genindex.html" title="General Index" accesskey="I">index</a></li> <li class="right" > <a href="../py-modindex.html" title="Python Module Index" >modules</a> |</li> <li class="right" > <a href="appendix.html" title="15. Appendix" accesskey="N">next</a> |</li> <li class="right" > <a href="interactive.html" title="13. Interactive Input Editing and History Substitution" accesskey="P">previous</a> |</li> <li><img src="../_static/py.png" alt="" style="vertical-align: middle; margin-top: -1px"/></li> <li><a href="https://www.python.org/">Python</a> »</li> <li> <a href="../index.html">Python 2.7.18 documentation</a> » </li> <li class="nav-item nav-item-1"><a href="index.html" accesskey="U">The Python Tutorial</a> »</li> </ul> </div> <div class="document"> <div class="documentwrapper"> <div class="bodywrapper"> <div class="body" role="main"> <div class="section" id="floating-point-arithmetic-issues-and-limitations"> <span id="tut-fp-issues"></span><h1>14. Floating Point Arithmetic: Issues and Limitations<a class="headerlink" href="#floating-point-arithmetic-issues-and-limitations" title="Permalink to this headline">¶</a></h1> <p>Floating-point numbers are represented in computer hardware as base 2 (binary) fractions. For example, the decimal fraction</p> <div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="mf">0.125</span> </pre></div> </div> <p>has value 1/10 + 2/100 + 5/1000, and in the same way the binary fraction</p> <div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="mf">0.001</span> </pre></div> </div> <p>has value 0/2 + 0/4 + 1/8. These two fractions have identical values, the only real difference being that the first is written in base 10 fractional notation, and the second in base 2.</p> <p>Unfortunately, most decimal fractions cannot be represented exactly as binary fractions. A consequence is that, in general, the decimal floating-point numbers you enter are only approximated by the binary floating-point numbers actually stored in the machine.</p> <p>The problem is easier to understand at first in base 10. Consider the fraction 1/3. You can approximate that as a base 10 fraction:</p> <div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="mf">0.3</span> </pre></div> </div> <p>or, better,</p> <div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="mf">0.33</span> </pre></div> </div> <p>or, better,</p> <div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="mf">0.333</span> </pre></div> </div> <p>and so on. No matter how many digits you’re willing to write down, the result will never be exactly 1/3, but will be an increasingly better approximation of 1/3.</p> <p>In the same way, no matter how many base 2 digits you’re willing to use, the decimal value 0.1 cannot be represented exactly as a base 2 fraction. In base 2, 1/10 is the infinitely repeating fraction</p> <div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="mf">0.0001100110011001100110011001100110011001100110011</span><span class="o">...</span> </pre></div> </div> <p>Stop at any finite number of bits, and you get an approximation.</p> <p>On a typical machine running Python, there are 53 bits of precision available for a Python float, so the value stored internally when you enter the decimal number <code class="docutils literal notranslate"><span class="pre">0.1</span></code> is the binary fraction</p> <div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="mf">0.00011001100110011001100110011001100110011001100110011010</span> </pre></div> </div> <p>which is close to, but not exactly equal to, 1/10.</p> <p>It’s easy to forget that the stored value is an approximation to the original decimal fraction, because of the way that floats are displayed at the interpreter prompt. Python only prints a decimal approximation to the true decimal value of the binary approximation stored by the machine. If Python were to print the true decimal value of the binary approximation stored for 0.1, it would have to display</p> <div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="gp">>>> </span><span class="mf">0.1</span> <span class="go">0.1000000000000000055511151231257827021181583404541015625</span> </pre></div> </div> <p>That is more digits than most people find useful, so Python keeps the number of digits manageable by displaying a rounded value instead</p> <div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="gp">>>> </span><span class="mf">0.1</span> <span class="go">0.1</span> </pre></div> </div> <p>It’s important to realize that this is, in a real sense, an illusion: the value in the machine is not exactly 1/10, you’re simply rounding the <em>display</em> of the true machine value. This fact becomes apparent as soon as you try to do arithmetic with these values</p> <div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="gp">>>> </span><span class="mf">0.1</span> <span class="o">+</span> <span class="mf">0.2</span> <span class="go">0.30000000000000004</span> </pre></div> </div> <p>Note that this is in the very nature of binary floating-point: this is not a bug in Python, and it is not a bug in your code either. You’ll see the same kind of thing in all languages that support your hardware’s floating-point arithmetic (although some languages may not <em>display</em> the difference by default, or in all output modes).</p> <p>Other surprises follow from this one. For example, if you try to round the value 2.675 to two decimal places, you get this</p> <div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="gp">>>> </span><span class="nb">round</span><span class="p">(</span><span class="mf">2.675</span><span class="p">,</span> <span class="mi">2</span><span class="p">)</span> <span class="go">2.67</span> </pre></div> </div> <p>The documentation for the built-in <a class="reference internal" href="../library/functions.html#round" title="round"><code class="xref py py-func docutils literal notranslate"><span class="pre">round()</span></code></a> function says that it rounds to the nearest value, rounding ties away from zero. Since the decimal fraction 2.675 is exactly halfway between 2.67 and 2.68, you might expect the result here to be (a binary approximation to) 2.68. It’s not, because when the decimal string <code class="docutils literal notranslate"><span class="pre">2.675</span></code> is converted to a binary floating-point number, it’s again replaced with a binary approximation, whose exact value is</p> <div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="mf">2.67499999999999982236431605997495353221893310546875</span> </pre></div> </div> <p>Since this approximation is slightly closer to 2.67 than to 2.68, it’s rounded down.</p> <p>If you’re in a situation where you care which way your decimal halfway-cases are rounded, you should consider using the <a class="reference internal" href="../library/decimal.html#module-decimal" title="decimal: Implementation of the General Decimal Arithmetic Specification."><code class="xref py py-mod docutils literal notranslate"><span class="pre">decimal</span></code></a> module. Incidentally, the <a class="reference internal" href="../library/decimal.html#module-decimal" title="decimal: Implementation of the General Decimal Arithmetic Specification."><code class="xref py py-mod docutils literal notranslate"><span class="pre">decimal</span></code></a> module also provides a nice way to “see” the exact value that’s stored in any particular Python float</p> <div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="gp">>>> </span><span class="kn">from</span> <span class="nn">decimal</span> <span class="k">import</span> <span class="n">Decimal</span> <span class="gp">>>> </span><span class="n">Decimal</span><span class="p">(</span><span class="mf">2.675</span><span class="p">)</span> <span class="go">Decimal('2.67499999999999982236431605997495353221893310546875')</span> </pre></div> </div> <p>Another consequence is that since 0.1 is not exactly 1/10, summing ten values of 0.1 may not yield exactly 1.0, either:</p> <div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="gp">>>> </span><span class="nb">sum</span> <span class="o">=</span> <span class="mf">0.0</span> <span class="gp">>>> </span><span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">10</span><span class="p">):</span> <span class="gp">... </span> <span class="nb">sum</span> <span class="o">+=</span> <span class="mf">0.1</span> <span class="gp">...</span> <span class="gp">>>> </span><span class="nb">sum</span> <span class="go">0.9999999999999999</span> </pre></div> </div> <p>Binary floating-point arithmetic holds many surprises like this. The problem with “0.1” is explained in precise detail below, in the “Representation Error” section. See <a class="reference external" href="http://www.lahey.com/float.htm">The Perils of Floating Point</a> for a more complete account of other common surprises.</p> <p>As that says near the end, “there are no easy answers.” Still, don’t be unduly wary of floating-point! The errors in Python float operations are inherited from the floating-point hardware, and on most machines are on the order of no more than 1 part in 2**53 per operation. That’s more than adequate for most tasks, but you do need to keep in mind that it’s not decimal arithmetic, and that every float operation can suffer a new rounding error.</p> <p>While pathological cases do exist, for most casual use of floating-point arithmetic you’ll see the result you expect in the end if you simply round the display of your final results to the number of decimal digits you expect. For fine control over how a float is displayed see the <a class="reference internal" href="../library/stdtypes.html#str.format" title="str.format"><code class="xref py py-meth docutils literal notranslate"><span class="pre">str.format()</span></code></a> method’s format specifiers in <a class="reference internal" href="../library/string.html#formatstrings"><span class="std std-ref">Format String Syntax</span></a>.</p> <div class="section" id="representation-error"> <span id="tut-fp-error"></span><h2>14.1. Representation Error<a class="headerlink" href="#representation-error" title="Permalink to this headline">¶</a></h2> <p>This section explains the “0.1” example in detail, and shows how you can perform an exact analysis of cases like this yourself. Basic familiarity with binary floating-point representation is assumed.</p> <p><em class="dfn">Representation error</em> refers to the fact that some (most, actually) decimal fractions cannot be represented exactly as binary (base 2) fractions. This is the chief reason why Python (or Perl, C, C++, Java, Fortran, and many others) often won’t display the exact decimal number you expect:</p> <div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="gp">>>> </span><span class="mf">0.1</span> <span class="o">+</span> <span class="mf">0.2</span> <span class="go">0.30000000000000004</span> </pre></div> </div> <p>Why is that? 1/10 and 2/10 are not exactly representable as a binary fraction. Almost all machines today (July 2010) use IEEE-754 floating point arithmetic, and almost all platforms map Python floats to IEEE-754 “double precision”. 754 doubles contain 53 bits of precision, so on input the computer strives to convert 0.1 to the closest fraction it can of the form <em>J</em>/2**<em>N</em> where <em>J</em> is an integer containing exactly 53 bits. Rewriting</p> <div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="mi">1</span> <span class="o">/</span> <span class="mi">10</span> <span class="o">~=</span> <span class="n">J</span> <span class="o">/</span> <span class="p">(</span><span class="mi">2</span><span class="o">**</span><span class="n">N</span><span class="p">)</span> </pre></div> </div> <p>as</p> <div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">J</span> <span class="o">~=</span> <span class="mi">2</span><span class="o">**</span><span class="n">N</span> <span class="o">/</span> <span class="mi">10</span> </pre></div> </div> <p>and recalling that <em>J</em> has exactly 53 bits (is <code class="docutils literal notranslate"><span class="pre">>=</span> <span class="pre">2**52</span></code> but <code class="docutils literal notranslate"><span class="pre"><</span> <span class="pre">2**53</span></code>), the best value for <em>N</em> is 56:</p> <div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="gp">>>> </span><span class="mi">2</span><span class="o">**</span><span class="mi">52</span> <span class="go">4503599627370496</span> <span class="gp">>>> </span><span class="mi">2</span><span class="o">**</span><span class="mi">53</span> <span class="go">9007199254740992</span> <span class="gp">>>> </span><span class="mi">2</span><span class="o">**</span><span class="mi">56</span><span class="o">/</span><span class="mi">10</span> <span class="go">7205759403792793</span> </pre></div> </div> <p>That is, 56 is the only value for <em>N</em> that leaves <em>J</em> with exactly 53 bits. The best possible value for <em>J</em> is then that quotient rounded:</p> <div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="gp">>>> </span><span class="n">q</span><span class="p">,</span> <span class="n">r</span> <span class="o">=</span> <span class="nb">divmod</span><span class="p">(</span><span class="mi">2</span><span class="o">**</span><span class="mi">56</span><span class="p">,</span> <span class="mi">10</span><span class="p">)</span> <span class="gp">>>> </span><span class="n">r</span> <span class="go">6</span> </pre></div> </div> <p>Since the remainder is more than half of 10, the best approximation is obtained by rounding up:</p> <div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="gp">>>> </span><span class="n">q</span><span class="o">+</span><span class="mi">1</span> <span class="go">7205759403792794</span> </pre></div> </div> <p>Therefore the best possible approximation to 1/10 in 754 double precision is that over 2**56, or</p> <div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="mi">7205759403792794</span> <span class="o">/</span> <span class="mi">72057594037927936</span> </pre></div> </div> <p>Note that since we rounded up, this is actually a little bit larger than 1/10; if we had not rounded up, the quotient would have been a little bit smaller than 1/10. But in no case can it be <em>exactly</em> 1/10!</p> <p>So the computer never “sees” 1/10: what it sees is the exact fraction given above, the best 754 double approximation it can get:</p> <div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="gp">>>> </span><span class="o">.</span><span class="mi">1</span> <span class="o">*</span> <span class="mi">2</span><span class="o">**</span><span class="mi">56</span> <span class="go">7205759403792794.0</span> </pre></div> </div> <p>If we multiply that fraction by 10**30, we can see the (truncated) value of its 30 most significant decimal digits:</p> <div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="gp">>>> </span><span class="mi">7205759403792794</span> <span class="o">*</span> <span class="mi">10</span><span class="o">**</span><span class="mi">30</span> <span class="o">//</span> <span class="mi">2</span><span class="o">**</span><span class="mi">56</span> <span class="go">100000000000000005551115123125L</span> </pre></div> </div> <p>meaning that the exact number stored in the computer is approximately equal to the decimal value 0.100000000000000005551115123125. In versions prior to Python 2.7 and Python 3.1, Python rounded this value to 17 significant digits, giving ‘0.10000000000000001’. In current versions, Python displays a value based on the shortest decimal fraction that rounds correctly back to the true binary value, resulting simply in ‘0.1’.</p> </div> </div> </div> </div> </div> <div class="sphinxsidebar" role="navigation" aria-label="main navigation"> <div class="sphinxsidebarwrapper"> <h3><a href="../contents.html">Table of Contents</a></h3> <ul> <li><a class="reference internal" href="#">14. Floating Point Arithmetic: Issues and Limitations</a><ul> <li><a class="reference internal" href="#representation-error">14.1. Representation Error</a></li> </ul> </li> </ul> <h4>Previous topic</h4> <p class="topless"><a href="interactive.html" title="previous chapter">13. Interactive Input Editing and History Substitution</a></p> <h4>Next topic</h4> <p class="topless"><a href="appendix.html" title="next chapter">15. 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