在 Pandas 中查找与数组匹配的列名 [英] Find column name in pandas that matches an array

问题描述

我有一个大数据框 (5000 x 12039)，我想获取与 numpy 数组匹配的列名.

I have a large dataframe (5000 x 12039) and I want to get the column name that matches a numpy array.

例如，如果我有桌子

        m1lenhr m1lenmin    m1citywt    m1a12a  cm1age  cm1numb m1b1a   m1b1b   m1b12a  m1b12b  ... kind_attention_scale_10 kind_attention_scale_22 kind_attention_scale_21 kind_attention_scale_15 kind_attention_scale_18 kind_attention_scale_19 kind_attention_scale_25 kind_attention_scale_24 kind_attention_scale_27 kind_attention_scale_23
challengeID                                                                                 
1   0.130765    40.0    202.485367  1.893256    27.0    1.0 2.0 0.0 2.254198    2.289966    ... 0   0   0   0   0   0   0   0   0   0
2   0.000000    40.0    45.608219   1.000000    24.0    1.0 2.0 0.0 2.000000    3.000000    ... 0   0   0   0   0   0   0   0   0   0
3   0.000000    35.0    39.060299   2.000000    23.0    1.0 2.0 0.0 2.254198    2.289966    ... 0   0   0   0   0   0   0   0   0   0
4   0.000000    30.0    22.304855   1.893256    22.0    1.0 3.0 0.0 2.000000    3.000000    ... 0   0   0   0   0   0   0   0   0   0
5   0.000000    25.0    35.518272   1.893256    19.0    1.0 1.0 6.0 1.000000    3.000000    ... 0

我想这样做:

x = [40.0, 40.0, 35.0, 30.0, 25.0]
find_column(x)

并让 find_column(x) 返回 m1lenmin

推荐答案

方法 #1

这是一种利用 NumPy 广播 -

Here's one vectorized approach leveraging NumPy broadcasting -

df.columns[(df.values == np.asarray(x)[:,None]).all(0)]

样品运行 -

In [367]: df
Out[367]: 
   0  1  2  3  4  5  6  7  8  9
0  7  1  2  6  2  1  7  2  0  6
1  5  4  3  3  2  1  1  1  5  5
2  7  7  2  2  5  4  6  6  5  7
3  0  5  4  1  5  7  8  2  2  4
4  7  1  0  4  5  4  3  2  8  6

In [368]: x = df.iloc[:,2].values.tolist()

In [369]: x
Out[369]: [2, 3, 2, 4, 0]

In [370]: df.columns[(df.values == np.asarray(x)[:,None]).all(0)]
Out[370]: Int64Index([2], dtype='int64')

方法#2

或者，这是另一个使用 views 概念的 -

Alternatively, here's another using the concept of views -

def view1D(a, b): # a, b are arrays
    a = np.ascontiguousarray(a)
    b = np.ascontiguousarray(b)
    void_dt = np.dtype((np.void, a.dtype.itemsize * a.shape[1]))
    return a.view(void_dt).ravel(),  b.view(void_dt).ravel()

df1D_arr, x1D = view1D(df.values.T,np.asarray(x)[None])
out = np.flatnonzero(df1D_arr==x1D)

样品运行 -

In [442]: df
Out[442]: 
   0  1  2  3  4  5  6  7  8  9
0  7  1  2  6  2  1  7  2  0  6
1  5  4  3  3  2  1  1  1  5  5
2  7  7  2  2  5  4  6  6  5  7
3  0  5  4  1  5  7  8  2  2  4
4  7  1  0  4  5  4  3  2  8  6

In [443]: x = df.iloc[:,5].values.tolist()

In [444]: df1D_arr, x1D = view1D(df.values.T,np.asarray(x)[None])

In [445]: np.flatnonzero(df1D_arr==x1D)
Out[445]: array([5])

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