[英]Delete certain elements of a numpy array
我有兩個 numpy arrays a
和b
。 我有一個定義,它構造了一個數組c
,其元素是 a 的不同元素的所有可能總和。
import numpy as np
def Sumarray(a):
n = len(a)
sumarray = np.array([0]) # Add a default zero element
for k in range(2,n+1):
full = np.mgrid[k*(slice(n),)]
nd_triu_idx = full[:,(np.diff(full,axis=0)>0).all(axis=0)]
sumarray = np.append(sumarray, a[nd_triu_idx].sum(axis=0))
return sumarray
a = np.array([1,2,6,8])
c = Sumarray(a)
print(d)
然后,我在c
的元素和b
之間執行子集和: isSubsetSum
返回b
的元素,求和后得到 c[1]。 假設我得到
c[0] = b[2] + b[3]
然后我想刪除:
a
的元素相加后得到 c[0] 從定義中可以看出, Sumarray
,保留了a
的不同元素之和的順序,所以我需要實現一些映射。
function isSubsetSum
由下式給出
def _isSubsetSum(numbers, n, x, indices):
if (x == 0):
return True
if (n == 0 and x != 0):
return False
# If last element is greater than x, then ignore it
if (numbers[n - 1] > x):
return _isSubsetSum(numbers, n - 1, x, indices)
# else, check if x can be obtained by any of the following
found = _isSubsetSum(numbers, n - 1, x, indices)
if found: return True
indices.insert(0, n - 1)
found = _isSubsetSum(numbers, n - 1, x - numbers[n - 1], indices)
if not found: indices.pop(0)
return found
def isSubsetSum(numbers, x):
indices = []
found = _isSubsetSum(numbers, len(numbers), x, indices)
return indices if found else None
當您迭代所有可能的術語數量時,您也可以直接生成所有可能的子集。
這些可以通過它們的二進制表示方便地編碼為數字 0,1,2,...:O 表示根本沒有項,1 表示只有第一項,2 表示只有第二項,3 表示第一個和第二個等等。
使用這種方案,從總和索引中恢復項變得非常容易,因為我們需要做的就是獲得二進制表示:
更新:我們可以用少量的額外代碼來抑制 1-term-sums:
import numpy as np
def find_all_subsums(a,drop_singletons=False):
n = len(a)
assert n<=32 # this gives 4G subsets, and we have to cut somewhere
# compute the smallest integer type with enough bits
dt = f"<u{1<<((n-1)>>3).bit_length()}"
# the numbers 0 to 2^n encode all possible subsets of an n
# element set by means of their binary representation
# each bit corresponds to one element number k represents the
# subset consisting of all elements whose bit is set in k
rng = np.arange(1<<n,dtype=dt)
if drop_singletons:
# one element subsets correspond to powers of two
rng = np.delete(rng,1<<np.arange(n))
# np.unpackbits transforms bytes to their binary representation
# given the a bitvector b we can compute the corresponding subsum
# as b dot a, to do it in bulk we can mutliply the matrix of
# binary rows with a
return np.unpackbits(rng[...,None].view('u1'),
axis=1,count=n,bitorder='little') @ a
def show_terms(a,idx,drop_singletons=False):
n = len(a)
if drop_singletons:
# we must undo the dropping of powers of two to get an index
# that is easy to translate. One can check that the following
# formula does the trick
idx += (idx+idx.bit_length()).bit_length()
# now we can simply use the binary representation
return a[np.unpackbits(np.asarray(idx,dtype='<u8')[None].view('u1'),
count=n,bitorder='little').view('?')]
example = np.logspace(1,7,7,base=3)
ss = find_all_subsums(example,True)
# check every single sum
for i,s in enumerate(ss):
assert show_terms(example,i,True).sum() == s
# print one example
idx = 77
print(ss[idx],"="," + ".join(show_terms(example.astype('U'),idx,True)))
樣品運行:
2457.0 = 27.0 + 243.0 + 2187.0
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