I want to remove rows from a ndarray based on another array. for example:
k = [1,3,99]
n = [
[1,'a']
[2,'b']
[3,'c']
[4,'c']
[.....]
[99, 'a']
[100,'e']
]
expect result:
out = [
[2,'b']
[4,'c']
[.....]
[100,'e']
]
the first column of the rows with the values in k will be removed
You can use np.in1d to create a mask of matches between the first column of n and k and then use the inverted mask to select the non-matching rows off n, like so -
n[~np.in1d(n[:,0].astype(int), k)]
If the first column is already of int dtype, skip the .astype(int) conversion step.
Sample run -
In [41]: n
Out[41]:
array([['1', 'a'],
['2', 'b'],
['3', 'c'],
['4', 'c'],
['99', 'a'],
['100', 'e']],
dtype='|S21')
In [42]: k
Out[42]: [1, 3, 99]
In [43]: n[~np.in1d(n[:,0].astype(int), k)]
Out[43]:
array([['2', 'b'],
['4', 'c'],
['100', 'e']],
dtype='|S21')
For peformance, if the first column is sorted, we can use np.searchsorted -
mask = np.ones(n.shape[0],dtype=bool)
mask[np.searchsorted(n[:,0], k)] = 0
out = n[mask]
n sorted? If it is, try with mask[np.searchsorted(n[:,0].astype(int), k)] = 0.If your data structure is list, please find simple solution as below, however you can convert into list by list() method.
def check(list):
k=[1,3,99]
if(list[0] not in k):
return list
final_list = map(check,n)
final_list = final_list.remove(None)
print final_list
