在列表中拆分基于 numpy 数组的列值

Question

I'am new in numpy and I want to split array 2D based on columns values if value in another list, I converted a pandas dataframe on numpy array of 2D and I have a another list, I want to split my numpy array on two others array, the first based on (if values of second column in list) and the second contains the rest of my numpy array, and I want to get the rest of my list(contains all values doesn't exist in my numpy array)

numpy_data = np.array([
        [1, 'p1', 2],
        [11, 'p2', 8],
        [1, 'p8', 21],
        [13, 'p10', 2] ])

list_value = ['p1', 'p3', 'p8']

The expected output :

data_in_list = [
        [1, 'p1', 2],
        [1, 'p8', 21]]
list_val_in_numpy = ['p1', 'p8'] # intersection of second column with my list

rest_data = [
        [11, 'p2', 8],
        [13, 'p10', 2]] 
rest_list_value = ['p3']

In my code I have found how to get first output :

first_output =  numpy_data[np.isin(numpy_data[:,1], list_value)]    

But I couldn't find the rest of my numpy, I have tried too, Browse my list and seek if values in second column of array and then delete this row, in this case I dont need the first output (That I called data_in_list, b-coz I do what I need on it), here I need the others output

for val in l :
    row = numpy_data[np.where(numpy_data[:,1]== val)]
    row.size != 0 :
        # My custom code
        # then remove this row from my numpy, I couldn't do it

Thanks in advance

Answer 1

在np.isin的结果上使用 python 的 invert ~运算符：

rest = numpy_data[~np.isin(numpy_data[:,1], list_value)]    

Answer 2

There are multiple ways of doing this. I would prefer a vectorized way of using list comprehension. But for sake of clarity here is loop way of doing the same thing.

data_in_list=[]
list_val_in_numpy = []
rest_data=[]
for x in numpy_data:
    for y in x:
        if y in list_value:
            data_in_list.append(x)
            for x in list_value:
                if x == y:
                    list_val_in_numpy.append(x)
for x in numpy_data:
    if x in data_in_list:
        pass
    else:
        rest_data.append(x)

This gives you all the three lists you were looking for. Concatenate to get the list you want exactly.

Answer 3

list comprehension will solve it I guess:

numpy_data = [
        [1, 'p1', 2],
        [11, 'p2', 8],
        [1, 'p8', 21],
        [13, 'p10', 2],

]

list_value = ['p1', 'p3', 'p8']

output_list = [[item] for item in numpy_data if item[1] in list_value]
print(output_list)

output:

[[[1, 'p1', 2]], [[1, 'p8', 21]]]