在Python中,可以获得lst1使用独有的元素:
lst1=['a','b','c']
lst2=['c','d','e']
lst3=[]
for i in lst1:
if i not in lst2:
lst3.append(i)
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什么是Matlab等价物?
Div*_*kar 14
您正在寻找MATLAB的setdiff-
setdiff(lst1,lst2)
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样品运行 -
>> lst1={'a','b','c'};
>> lst2={'c','d','e'};
>> setdiff(lst1,lst2)
ans =
'a' 'b'
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用Python运行验证 -
In [161]: lst1=['a','b','c']
...: lst2=['c','d','e']
...: lst3=[]
...: for i in lst1:
...: if i not in lst2:
...: lst3.append(i)
...:
In [162]: lst3
Out[162]: ['a', 'b']
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事实上,你setdiff在Python NumPy module中也是如此numpy.setdiff1d.与它等效的实现将是 -
In [166]: import numpy as np
In [167]: np.setdiff1d(lst1,lst2) # Output as an array
Out[167]:
array(['a', 'b'],
dtype='|S1')
In [168]: np.setdiff1d(lst1,lst2).tolist() # Output as list
Out[168]: ['a', 'b']
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