Python 如何计算sem（）？

首先导入熊猫并创建具有完美正态分布的系列：

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，分布是否正常：

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

现在让我们打印宇宙的sem（）：

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

现在查看示例：

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

但我不明白熊猫如何计算标准误差的平均值，若它和宇宙。它有用吗

se_x = sd_x / sqrt(len(x))

还是创建样本？如果它创建了样本，我可以设置它们的数量和方式

如果计数<30，熊猫如何计算样本的sem？

:

您可能还希望检查模块中可用的方法

    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)

@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)