本文介绍了在 Pandas 中查找与数组匹配的列名的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
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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