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python - Python：如何从矩阵中迭代地删除行和列？

问题描述

我是初学者，自学成才。

我想为方阵迭代地删除行和列。

我的第一次尝试是：

for i in range(0,c):
    if Atomos[i] == "H":
        d_nh=np.delete(np.delete(d, i, 0), i, 1)

我的来源

但是最终的和新的矩阵显示只删除了一行和一列。

如何迭代地删除行和列？

这两个文件的内容是：

马特里兹

0.000000000000000000e+00   1.205475107329385898e+00   1.342931901022796248e+00   2.343013632351988562e+00   3.227383136403330077e+00   2.640130058756467779e+00   2.640148435557436279e+00   1.478495377186577864e+00   2.442752671162299549e+00   3.440470104931585560e+00   2.650641510969556247e+00   2.173942147030341054e+00
1.205475107329385898e+00   0.000000000000000000e+00   2.245467917547002035e+00   2.644315603095303224e+00   3.702905546101439072e+00   2.635453659417908323e+00   2.635572456117051487e+00   2.391886453689349601e+00   2.871975783234887114e+00   3.947951548910517250e+00   2.593644960074543704e+00   3.289694675733229268e+00
1.342931901022796248e+00   2.245467917547002035e+00   0.000000000000000000e+00   1.418915551312474932e+00   2.015476882415431881e+00   2.069308813492318766e+00   2.069295883966994598e+00   2.323619373652348497e+00   3.560975969980455957e+00   4.431347320573396864e+00   3.951843753512012114e+00   2.436642114389359737e+00
2.343013632351988562e+00   2.644315603095303224e+00   1.418915551312474932e+00   0.000000000000000000e+00   1.086884605635873857e+00   1.092126176004005522e+00   1.092126228351472994e+00   3.641924623709103415e+00   4.772348473634059118e+00   5.725281935435472036e+00   4.948741644534886852e+00   3.855293676517856927e+00
3.227383136403330077e+00   3.702905546101439072e+00   2.015476882415431881e+00   1.086884605635873857e+00   0.000000000000000000e+00   1.791611832133639171e+00   1.791607398071044743e+00   4.336840746006843439e+00   5.570012200282658199e+00   6.444369626625309877e+00   5.876935928592362579e+00   4.304036910039308772e+00
2.640130058756467779e+00   2.635453659417908323e+00   2.069308813492318766e+00   1.092126176004005522e+00   1.791611832133639171e+00   0.000000000000000000e+00   1.774322615322815988e+00   3.999843247699306215e+00   5.001451201004137204e+00   5.992370839831868246e+00   5.038926795069470721e+00   4.349546588337785735e+00
2.640148435557436279e+00   2.635572456117051487e+00   2.069295883966994598e+00   1.092126228351472994e+00   1.791607398071044743e+00   1.774322615322815988e+00   0.000000000000000000e+00   3.999902964280430240e+00   5.001556219427222061e+00   5.992449776200326816e+00   5.039085741282741004e+00   4.349558376763067891e+00
1.478495377186577864e+00   2.391886453689349601e+00   2.323619373652348497e+00   3.641924623709103415e+00   4.336840746006843439e+00   3.999843247699306215e+00   3.999902964280430240e+00   0.000000000000000000e+00   1.324770443414402976e+00   2.107792016824585168e+00   2.085364895492880954e+00   1.079295724832157077e+00
2.442752671162299549e+00   2.871975783234887114e+00   3.560975969980455957e+00   4.772348473634059118e+00   5.570012200282658199e+00   5.001451201004137204e+00   5.001556219427222061e+00   1.324770443414402976e+00   0.000000000000000000e+00   1.076370750308789148e+00   1.078101361047288487e+00   2.119237286319515245e+00
3.440470104931585560e+00   3.947951548910517250e+00   4.431347320573396864e+00   5.725281935435472036e+00   6.444369626625309877e+00   5.992370839831868246e+00   5.992449776200326816e+00   2.107792016824585168e+00   1.076370750308789148e+00   0.000000000000000000e+00   1.841888017015948842e+00   2.494970001809259763e+00
2.650641510969556247e+00   2.593644960074543704e+00   3.951843753512012114e+00   4.948741644534886852e+00   5.876935928592362579e+00   5.038926795069470721e+00   5.039085741282741004e+00   2.085364895492880954e+00   1.078101361047288487e+00   1.841888017015948842e+00   0.000000000000000000e+00   3.067298402780731070e+00
2.173942147030341054e+00   3.289694675733229268e+00   2.436642114389359737e+00   3.855293676517856927e+00   4.304036910039308772e+00   4.349546588337785735e+00   4.349558376763067891e+00   1.079295724832157077e+00   2.119237286319515245e+00   2.494970001809259763e+00   3.067298402780731070e+00   0.000000000000000000e+00

原子弹

C
O
O
C
H
H
H
C
C
H
H
H

标签： python

解决方案

我刚刚回答了你之前的问题。既然我完全理解了这个问题，我认为你有更好的方法来解决你的问题。

如果您要从 Atomos 包含“H”的“Matriz”中删除行。您应该将 2 个数据集合并为一个，并根据哪些行包含“H”进行过滤

编辑：更新响应以通过存储在 Atomos 中的列名反映 Matriz 中的过滤（列和行）

Matriz = pd.read_csv('test1.txt', delim_whitespace=True, header=None)
Atomos = pd.read_csv('test.txt',names=['index'])
columns = Atomos['index'].to_list()
Matriz.columns = columns
combined = pd.concat([Atomos,Matriz], axis=1)
combined[['C','O']][combined['index'] == 'H']

更新的输出


C   C   C   C   O   O
4   3.227383    1.086885    4.336841    5.570012    3.702906    2.015477
5   2.640130    1.092126    3.999843    5.001451    2.635454    2.069309
6   2.640148    1.092126    3.999903    5.001556    2.635572    2.069296
9   3.440470    5.725282    2.107792    1.076371    3.947952    4.431347
10  2.650642    4.948742    2.085365    1.078101    2.593645    3.951844
11  2.173942    3.855294    1.079296    2.119237    3.289695    2.436642

这会快得多。让我知道这是如何工作的。

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