Question

首先导入pandas并创建具有完美正态分布的Series：

import pandas as pd

lst = [[5 for x in range(5)], [4 for x in range(4)], [3 for x in range(3)],
       [2 for x in range(2)], [1 for x in range(1)], [2 for x in range(2)],
       [3 for x in range(3)], [4 for x in range(4)], [5 for x in range(5)]]

lst = [item for sublists in lst for item in sublists]

series = pd.Series(lst)

让我们检查一下，这种分布是正常的：

print(round(sum(series - series.mean()) / series.count(), 1) == 0)
# if distribution is normal we'll see True

现在让我们为宇宙打印sem（）：

print(series.sem(ddof=0))
# 0.21619987017

现在来样品：

print(series.sem()) # ddof=1
# 0.220026713637

但是我无法理解如果它与宇宙一起工作，熊猫如何计算平均值的标准误差。是否使用

se_x = sd_x / sqrt(len(x))

还是创建样品？如果它创建样本，我可以设置多少以及如何设置它们的数量？

如果计数＆lt; pandas如何计算样本的sem 30？

Answer 1

Pandas generates sem method dynamically

    cls.sem = _make_stat_function_ddof(
        cls, 'sem', name, name2, axis_descr,
        "Return unbiased standard error of the mean over requested "
        "axis.\n\nNormalized by N-1 by default. This can be changed "
        "using the ddof argument",
        nanops.nansem)

where nanops.nansem() is：

@disallow('M8', 'm8')
def nansem(values, axis=None, skipna=True, ddof=1):
    var = nanvar(values, axis, skipna, ddof=ddof)

    mask = isnull(values)
    if not is_float_dtype(values.dtype):
        values = values.astype('f8')
    count, _ = _get_counts_nanvar(mask, axis, ddof, values.dtype)
    var = nanvar(values, axis, skipna, ddof=ddof)

    return np.sqrt(var) / np.sqrt(count)

您可能还想查看scipy.stats模块

中可用的方法

pandas如何计算sem（）？

1 个答案: