我有一个大型数据帧(5000 x 12039),我想获得与numpy数组匹配的列名.
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
我想做这个：
x = [40.0,40.0,35.0,30.0,25.0]
find_column(x)
并让find_column(x)返回m1lenmin
最佳答案
方法#1
原文链接：https://www.f2er.com/python/438736.html这是一个利用NumPy broadcasting的矢量化方法 –
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,这是另一个使用观点的概念 –
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)
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,np.asarray(x)[None])
In [445]: np.flatnonzero(df1D_arr==x1D)
Out[445]: array([5])