在pandas中查找与数组匹配的列名

时间:2017-07-25 20:53:23

标签: python pandas numpy dataframe

我有一个大型数据框(5000 x 12039),我想获得与numpy数组匹配的列名。

例如,如果我有表

        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

2 个答案:

答案 0 :(得分:6)

方法#1

这是一种利用NumPy broadcasting -

的矢量化方法
df.columns[(df.values == np.asarray(x)[:,None]).all(0)]

示例运行 -

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 -

的概念
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)

示例运行 -

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])

答案 1 :(得分:5)

试试这个:

In [91]: x = np.array(x)

In [94]: df.apply(lambda col: col.eq(x).all())
Out[94]:
m1lenhr     False
m1lenmin     True
m1citywt    False
m1a12a      False
cm1age      False
cm1numb     False
m1b1a       False
m1b1b       False
m1b12a      False
m1b12b      False
dtype: bool

In [95]: df.columns[df.apply(lambda col: col.eq(x).all()).values]
Out[95]: Index(['m1lenmin'], dtype='object')