pandas 在多索引上应用函数 [英] pandas apply function on multiindex

问题描述

我想在多索引数据帧(基本上是 groupby 描述数据帧)上应用一个函数，而不使用 for 循环来遍历 0 级索引.

I would like to apply a function on a multiindex dataframe (basically groupby describe dataframe) without using for loop to traverse level 0 index.

我想申请的职能:

def CI(x):
    import math
    sigma = x["std"]
    n = x["count"]
    return 1.96 * sigma / math.sqrt(n)

我的数据框示例:

df = df.iloc[47:52, [3,4,-1]]

               a          b                    id
47          0.218182   0.000000  0d1974107c6731989c762e96def73568
48          0.000000   0.000000  0d1974107c6731989c762e96def73568
49          0.218182   0.130909  0d1974107c6731989c762e96def73568
50          0.000000   0.000000  0fd4f3b4adf43682f08e693a905b7432
51          0.000000   0.000000  0fd4f3b4adf43682f08e693a905b7432

我用 nan 替换零:

And I replace zeros with nan:

df = df.replace(float(0), np.nan)

Groupy on id 和 describe，我得到多索引:

Groupy on id and describe and I get multiindex:

df_group = df.groupby("id").describe()

我不喜欢当前的解决方案，我认为可以改进:

Current solution I don't like and think could be improved:

l_df = []
for column in df_group.columns.levels[0]:
    df = pd.DataFrame({"CI" : df_group[column].apply(CI, axis = 1)})
    l_df.append(df)
CI = pd.concat(l_df, axis = 1)
CI.columns = df_group.columns.levels[0]

所以我得到类似的信息:

so I get something like:

                                    a       b
id
06f32e6e45da385834dac983256d59f3    nan     nan
0d1974107c6731989c762e96def73568    0.005   0.225
0fd4f3b4adf43682f08e693a905b7432    0.008   nan
11e0057cdc8b8e1b1cdabfa8a092ea5f    0.018   0.582
120549af6977623bd01d77135a91a523    0.008   0.204

同样，如果我有从 a 到 z 的顶级列，并且每个列都包含 std 和 count 列，我如何同时将我的函数应用于这些列中的每一个?

So again, if I have top level columns from a to z, and each contains std and count column, how can I apply my function to each of these columns at the same time?

推荐答案

在 level 上使用 groupby 和 axis=1，让你迭代并应用于第一级列.

Using groupby on level with axis=1, let's you iterate and apply over the first level columns.

In [104]: (df.groupby("id").describe()
             .groupby(level=0, axis=1)
             .apply(lambda x: x[x.name].apply(CI, axis=1)))
Out[104]:
                                    a   b
id
0d1974107c6731989c762e96def73568  0.0 NaN
0fd4f3b4adf43682f08e693a905b7432  NaN NaN

事实上，你不需要CI，如果你想

Infact, you don't need CI, if you were to

In [105]: (df.groupby("id").describe()
             .groupby(level=0, axis=1).apply(lambda x: x[x.name]
             .apply(lambda x: 1.96*x['std']/np.sqrt(x['count']), axis=1)))
Out[105]:
                                    a   b
id
0d1974107c6731989c762e96def73568  0.0 NaN
0fd4f3b4adf43682f08e693a905b7432  NaN NaN

示例 df

In [106]: df
Out[106]:
           a         b                                id
47  0.218182       NaN  0d1974107c6731989c762e96def73568
48       NaN       NaN  0d1974107c6731989c762e96def73568
49  0.218182  0.130909  0d1974107c6731989c762e96def73568
50       NaN       NaN  0fd4f3b4adf43682f08e693a905b7432
51       NaN       NaN  0fd4f3b4adf43682f08e693a905b7432

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