Question

import scipy

timeseries = [53.0, 28.0, 20.0, 113.0, 68.0, 18.0, 9.0, 72.0, 37.0, 29.0, 16.0, 70.0, 45.0, 3.0, 79.0, 7.0, 17.0, 0.0, 84.0, 19.0,
          0.0, 1.0, 5.0, 16.0, 1485.3333, 650.0, 39.0, 52.0, 82.0, 13.0, 11.0, 14.0, 31.0, 20.0, 399.0, 124.0, 39.0, 0.0, 9.0,
          42.0, 41.0, 98.5, 10.0, 4.0, 19.0, 53.0, 60.0, 789.0, 1471.3333, 876.0, 5.0, 714.0, 136.0, 27.0, 38.0, 29.0, 10.0,
          181.0, 1.0, 14.0, 39.0, 29.0, 2.0, 1502.0, 174.5, 4.0, 305.0, 222.6667, 349.0, 38.0, 15.0, 168.0, 41.0, 28.0, 15.0,
          508.0, 57.0, 26.0, 146.0, 50.5, 20.0, 12.0, 10.0, 15.0, 3.0, 19.0, 2922.0, 5200.5, 2989.0, 0.0, 5.0, 13.0, 2.0, 2.0,
          4.0, 32.0, 66.0, 4.0, 36.0, 1.0, 6.0, 8.0, 88.0, 3.0, 7.0, 250.0, 0.5, 9.0, 0.0, 94.0, 16.0, 3.0, 6.0, 15.0, 4.0, 4.0,
          240.0, 266.6667, 1208.0, 2387.0, 3883.5, 2997.3333, 2667.0, 417.5, 3.0, 26.0, 15.0, 11.0, 4.0, 70.0, 202.0, 2.0, 13.0,
          3.0, 1.0, 6.0, 7.0, 5.0, 140.0, 954.0, 2343.0, 5264.6667, 6051.5, 1181.0, 489.5, 879.0, 1531.0, 2064.3333, 1472.0,
          2029.3333, 3112.0, 2232.6667, 45.0, 716.5, 997.0, 1374.6667, 1993.5, 2549.0, 2690.5, 2640.3333, 2514.5, 1230.0, 475.5,
          893.3333, 1984.5, 2054.3333, 1800.5, 2793.3333, 3630.5, 4305.3333, 5214.0, 5790.6667]

series = scipy.array(timeseries)
stdDev = scipy.std(series, dtype=scipy.float64)

print stdDev

返回1246.16323355，而来自Commons Math的Java程序返回1249.801674091763
如果我用http://easycalculation.com/statistics/standard-deviation.php检查它，它也会返回1249.80167

Scipy标准偏差有什么问题？

Answer 1

阅读numpy.std的文档字符串的Notes部分（与scipy.std相同）。默认情况下，std将平方偏差之和除以n。要获得与您使用ddof=1的其他工具返回的值相匹配的值，以使其除以n - 1：

In [2]: a = np.array(timeseries)

In [3]: std(a)
Out[3]: 1246.1632335502143

In [4]: std(a, ddof=1)
Out[4]: 1249.8016740917631

为什么Scipy stdDev会返回错误的结果？

1 个答案: