问题
Is there a simple way to index all elements of a list (or array, or whatever) except for a particular index? E.g.,
mylist[3]will return the item in position 3milist[~3]will return the whole list except for 3
回答1:
For a list, you could use a list comp. For example, to make b a copy of a without the 3rd element:
a = range(10)[::-1] # [9, 8, 7, 6, 5, 4, 3, 2, 1, 0]
b = [x for i,x in enumerate(a) if i!=3] # [9, 8, 7, 5, 4, 3, 2, 1, 0]
This is very general, and can be used with all iterables, including numpy arrays. If you replace [] with (), b will be an iterator instead of a list.
Or you could do this in-place with pop:
a = range(10)[::-1] # a = [9, 8, 7, 6, 5, 4, 3, 2, 1, 0]
a.pop(3) # a = [9, 8, 7, 5, 4, 3, 2, 1, 0]
In numpy you could do this with a boolean indexing:
a = np.arange(9, -1, -1) # a = array([9, 8, 7, 6, 5, 4, 3, 2, 1, 0])
b = a[np.arange(len(a))!=3] # b = array([9, 8, 7, 5, 4, 3, 2, 1, 0])
which will, in general, be much faster than the list comprehension listed above.
回答2:
>>> l = range(1,10)
>>> l
[1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> l[:2]
[1, 2]
>>> l[3:]
[4, 5, 6, 7, 8, 9]
>>> l[:2] + l[3:]
[1, 2, 4, 5, 6, 7, 8, 9]
>>>
See also
Explain Python's slice notation
回答3:
The simplest way I found was:
mylist[:x]+mylist[x+1:]
that will produce your mylist without the element at index x.
回答4:
If you are using numpy, the closest, I can think of is using a mask
>>> import numpy as np
>>> arr = np.arange(1,10)
>>> mask = np.ones(arr.shape,dtype=bool)
>>> mask[5]=0
>>> arr[mask]
array([1, 2, 3, 4, 5, 7, 8, 9])
Something similar can be achieved using itertools without numpy
>>> from itertools import compress
>>> arr = range(1,10)
>>> mask = [1]*len(arr)
>>> mask[5]=0
>>> list(compress(arr,mask))
[1, 2, 3, 4, 5, 7, 8, 9]
回答5:
Use np.delete ! It does not actually delete anything inplace
Example:
import numpy as np
a = np.array([[1,4],[5,7],[3,1]])
# a: array([[1, 4],
# [5, 7],
# [3, 1]])
ind = np.array([0,1])
# ind: array([0, 1])
# a[ind]: array([[1, 4],
# [5, 7]])
all_except_index = np.delete(a, ind, axis=0)
# all_except_index: array([[3, 1]])
# a: (still the same): array([[1, 4],
# [5, 7],
# [3, 1]])
回答6:
If you don't know the index beforehand here is a function that will work
def reverse_index(l, index):
try:
l.pop(index)
return l
except IndexError:
return False
来源:https://stackoverflow.com/questions/19286657/index-all-except-one-item-in-python