Question

我正在尝试降低以下问题的速度性能。我有一个数组，例如：

list1 = [0.564,0.011,0.560,-1.100,0.344,0.912,-0.983]
list2 = [0.0,1.0,1.0,0.0,0.0,0.0,-1.0]

list3 = [0.760,0.013,-0.580,1.120,0.144,-0.929,0.833]
list4 = [-1.0,1.0,0.0,1.0,0.0,0.0,1.0]

test_arr = np.column_stack((list1, list2,list3,list4))

给出：

我总是会有一列不同的浮点数（让我们将这些列称为“ random_numbers”），然后是另一列，仅包含-1.0、0.0和1.0值的混合（让我们将这些列称为“ ones_zeros”）。

最终目标是将任何-1.0或1.0（注意：不是0.0）值替换为紧靠左侧的值。对于此示例，输出为：

当前，我正在将numpy数组转换为pandas并应用以下功能：

def replace_values(test_arr_df,random_numbers,ones_zeros):

    for cc in range(len(random_numbers)):

        test_arr_df[ones_zeros[cc]] = test_arr_df.apply(
            lambda row: row[random_numbers[cc]] if row[ones_zeros[cc]]==1 or row[ones_zeros[cc]]==-1
            else row[ones_zeros[cc]],axis=1

        )

    return test_arr_df

将其应用于我们的测试用例：

#Convert to dataframe
test_arr_df=pd.DataFrame(test_arr)

#Tell the function what is a variable column and what is a minmax column
variable_columns = [0,2]; minmax_columns = [1,3]

#Replace values
res_df = replace_values(test_arr_df,variable_columns,minmax_columns)

此pandas方法有效，其结果与上面的示例输出相同。但是，它非常慢。在代码的其他部分，我通过保留numpy数组而不切换到熊猫来成功地减少了处理时间，但是在这里我没有成功。

所以，我的问题是，有没有办法使用numpy而不是pandas来做到这一点？还是使用熊猫的更快方法？我无法取得进展，因为我一直在索引错误的部分或无法替换正确的行/列。谢谢！

Answer 1

您可以使用np.where替换值：

import numpy as np
import pandas as pd

list1 = [0.564,0.011,0.560,-1.100,0.344,0.912,-0.983]
list2 = [0.0,1.0,1.0,0.0,0.0,0.0,-1.0]

list3 = [0.760,0.013,-0.580,1.120,0.144,-0.929,0.833]
list4 = [-1.0,1.0,0.0,1.0,0.0,0.0,1.0]

df = pd.DataFrame({0:list1, 1:list2, 2:list3, 3:list4})

df.iloc[:, 1::2] = np.where(df.iloc[:, 1::2].isin([1, -1]), df.iloc[:, ::2], 0)
print(df.to_numpy())

打印：

[[ 0.564  0.     0.76   0.76 ]
 [ 0.011  0.011  0.013  0.013]
 [ 0.56   0.56  -0.58   0.   ]
 [-1.1    0.     1.12   1.12 ]
 [ 0.344  0.     0.144  0.   ]
 [ 0.912  0.    -0.929  0.   ]
 [-0.983 -0.983  0.833  0.833]]

编辑：版本，其中明确选择了列名称：

import numpy as np
import pandas as pd

list1 = [0.564,0.011,0.560,-1.100,0.344,0.912,-0.983]
list2 = [0.0,1.0,1.0,0.0,0.0,0.0,-1.0]

list3 = [0.760,0.013,-0.580,1.120,0.144,-0.929,0.833]
list4 = [-1.0,1.0,0.0,1.0,0.0,0.0,1.0]

df = pd.DataFrame({'Pressure':list1, 'Pressure 0-1':list2, 'Temperature':list3, 'Temperature 0-1':list4})

df[['Pressure 0-1', 'Temperature 0-1']] = np.where(df[['Pressure 0-1', 'Temperature 0-1']].isin([1, -1]), df[ ['Pressure', 'Temperature'] ], 0)
print(df)

打印：

   Pressure  Pressure 0-1  Temperature  Temperature 0-1
0     0.564         0.000        0.760            0.760
1     0.011         0.011        0.013            0.013
2     0.560         0.560       -0.580            0.000
3    -1.100         0.000        1.120            1.120
4     0.344         0.000        0.144            0.000
5     0.912         0.000       -0.929            0.000
6    -0.983        -0.983        0.833            0.833

Answer 2

这里：

for x, y in np.argwhere(np.abs(test_arr) == 1.):
    test_arr[x, y] = test_arr[x, y-1]

之前：

[[ 0.564  0.     0.76  -1.   ]
 [ 0.011  1.     0.013  1.   ]
 [ 0.56   1.    -0.58   0.   ]
 [-1.1    0.     1.12   1.   ]
 [ 0.344  0.     0.144  0.   ]
 [ 0.912  0.    -0.929  0.   ]
 [-0.983 -1.     0.833  1.   ]]

之后：

[[ 0.564  0.     0.76   0.76 ]
 [ 0.011  0.011  0.013  0.013]
 [ 0.56   0.56  -0.58   0.   ]
 [-1.1    0.     1.12   1.12 ]
 [ 0.344  0.     0.144  0.   ]
 [ 0.912  0.    -0.929  0.   ]
 [-0.983 -0.983  0.833  0.833]]

逻辑：对于值x或y的所有1和-1坐标，用左侧的值替换。

将数组中特定列中的值替换为相邻列中的值

2 个答案: