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问题:
Let's say we have a 1d numpy array filled with some int values. And let's say that some of them are 0.
Is there any way, using numpy array's power, to fill all the 0 values with the last non-zero values found?
for example:
arr = np.array([1, 0, 0, 2, 0, 4, 6, 8, 0, 0, 0, 0, 2]) fill_zeros_with_last(arr) print arr [1 1 1 2 2 4 6 8 8 8 8 8 2]
A way to do it would be with this function:
def fill_zeros_with_last(arr): last_val = None # I don't really care about the initial value for i in range(arr.size): if arr[i]: last_val = arr[i] elif last_val is not None: arr[i] = last_val
However, this is using a raw python for loop instead of taking advantage of the numpy and scipy power.
If we knew that a reasonably small number of consecutive zeros are possible, we could use something based on numpy.roll. The problem is that the number of consecutive zeros is potentially large...
Any ideas? or should we go straight to Cython?
Disclaimer:
I would say long ago I found a question in stackoverflow asking something like this or very similar. I wasn't able to find it. :-(
Maybe I missed the right search terms, sorry for the duplicate then. Maybe it was just my imagination...
回答1:
Here's a solution using np.maximum.accumulate:
def fill_zeros_with_last(arr): prev = np.arange(len(arr)) prev[arr == 0] = 0 prev = np.maximum.accumulate(prev) return arr[prev]
We construct an array prev which has the same length as arr, and such that prev[i] is the index of the last non-zero entry before the i-th entry of arr. For example, if:
>>> arr = np.array([1, 0, 0, 2, 0, 4, 6, 8, 0, 0, 0, 0, 2])
Then prev looks like:
array([ 0, 0, 0, 3, 3, 5, 6, 7, 7, 7, 7, 7, 12])
Then we just index into arr with prev and we obtain our result. A test:
>>> arr = np.array([1, 0, 0, 2, 0, 4, 6, 8, 0, 0, 0, 0, 2]) >>> fill_zeros_with_last(arr) array([1, 1, 1, 2, 2, 4, 6, 8, 8, 8, 8, 8, 2])
Note: Be careful to understand what this does when the first entry of your array is zero:
>>> fill_zeros_with_last(np.array([0,0,1,0,0])) array([0, 0, 1, 1, 1])
回答2:
Inspired by jme's answer here and by Bas Swinckels' (in the linked question) I came up with a different combination of numpy functions:
def fill_zeros_with_last(arr, initial=0): ind = np.nonzero(arr)[0] cnt = np.cumsum(np.array(arr, dtype=bool)) return np.where(cnt, arr[ind[cnt-1]], initial)
I think it's succinct and also works, so I'm posting it here for the record. Still, jme's is also succinct and easy to follow and seems to be faster, so I'm accepting it :-)
回答3:
If the 0s only come in strings of 1, this use of nonzero might work:
In [266]: arr=np.array([1,0,2,3,0,4,0,5]) In [267]: I=np.nonzero(arr==0)[0] In [268]: arr[I] = arr[I-1] In [269]: arr Out[269]: array([1, 1, 2, 3, 3, 4, 4, 5])
I can handle your arr by applying this repeatedly until I is empty.
In [286]: arr = np.array([1, 0, 0, 2, 0, 4, 6, 8, 0, 0, 0, 0, 2]) In [287]: while True: .....: I=np.nonzero(arr==0)[0] .....: if len(I)==0: break .....: arr[I] = arr[I-1] .....: In [288]: arr Out[288]: array([1, 1, 1, 2, 2, 4, 6, 8, 8, 8, 8, 8, 2])
If the strings of 0s are long it might be better to look for those strings and handle them as a block. But if most strings are short, this repeated application may be the fastest route.
