使用numpy
或itertools
是确定到下一个非连续元素的距离的有效方法。
> import numpy as np
> a=np.array(['a','b','b','c','d','a','b','b','c','c','c','d'])
我希望输出为。
[1,2,1,1,1,1,2,1,3,2,1]
扩展这个,我想要两个新元素的距离。预期的输出应该是
[3,3,2,2,2,3,5,4]
作为a
之后的两个新元素是b
(两个)和c
,依此类推。
修改1 我有两个版本用于查找下一个新元素:
import numpy as np
a = np.array(['a', 'b', 'b', 'c', 'd', 'a', 'b', 'b', 'c', 'c', 'c', 'd'])
# Using numpy
u, idx = np.unique(a, return_inverse=True)
idx = np.diff(idx)
idx[idx < 0] = 1
idx[idx > 1] = 1
count = 1
while 0 in idx:
idx[np.diff(idx) == count] = count+1
count += 1 │
print idx
# Using loop
oldElement = a[0]
dist = []
count = 1
for elm in a[1:]:
if elm == oldElement:
count += 1
else:
dist.extend(range(count, 0, -1))
count = 1
oldElement = elm
print dist
然而,这种方法不能简单地扩展到找到2个新元素。
答案 0 :(得分:1)
不幸的是,我没有针对一般问题的numpy / vectorized解决方案。
以下是一般解决方案,适用于任何深度。问题的第一部分对应于 depth = 1 ,第二部分对应于 depth = 2 。此解决方案也适用于更高的深度。
显然,如果您只想解决 depth = 1 的情况,可以提出一个更简单的解决方案。但是,对于这个问题,通用性增加了复杂性。
from itertools import groupby, chain, izip
ilen = lambda it: sum(1 for dummy in it)
def get_squeezed_counts(a):
"""
squeeze a sequence to a sequnce of value/count.
E.g. ['a', 'a', 'a', 'b'] --> [['a',3], ['b',1]]
"""
return [ [ v, ilen(it) ] for v, it in groupby(a) ]
def get_element_dist(counts, index, depth):
"""
For a given index in a "squeezed" list, return the distance (in the
original-array) with a given depth, or None.
E.g.
get_element_dist([['a',1],['b',2],['c',1]], 0, depth=1) --> 1 # from a to first b
get_element_dist([['a',1],['b',2],['c',1]], 1, depth=1) --> 2 # from first b to c
get_element_dist([['a',1],['b',2],['c',1]], 0, depth=2) --> 3 # from a to c
get_element_dist([['a',1],['b',2],['c',1]], 1, depth=2) --> None # from first b to end of sequence
"""
seen = set()
sum_counts = 0
for i in xrange(index, len(counts)):
v, count = counts[i]
seen.add(v)
if len(seen) > depth:
return sum_counts
sum_counts += count
# reached end of sequence before finding the next value
return None
def get_squeezed_dists(counts, depth):
"""
Construct a per-squeezed-element distance list, by calling get_element_dist()
for each element in counts.
E.g.
get_squeezed_dists([['a',1],['b',2],['c',1]], depth=1) --> [1,2,None]
"""
return [ get_element_dist(counts, i, depth=depth) for i in xrange(len(counts)) ]
def get_dists(a, depth):
counts = get_squeezed_counts(a)
squeezed_dists = get_squeezed_dists(counts, depth=depth)
# "Unpack" squeezed dists:
return list(chain.from_iterable(
xrange(dist, dist-count, -1)
for (v, count), dist in izip(counts, squeezed_dists)
if dist is not None
))
print get_dists(['a','b','b','c','d','a','b','b','c','c','c','d'], depth = 1)
# => [1, 2, 1, 1, 1, 1, 2, 1, 3, 2, 1]
print get_dists(['a','a','a'], depth = 1)
# => []
print get_dists(['a','b','b','c','d','a','b','b','c','c','c','d'], depth = 2)
# => [3, 3, 2, 2, 2, 3, 5, 4]
print get_dists(['a','b','a', 'b'], depth = 2)
# => []
对于python3,请替换xrange->range
和izip->zip
。
答案 1 :(得分:0)
这是我尝试距离一个元素。
import numpy as np
a=np.array(['a','b','b','c','d','a','b','b','c','c','c','d'])
out = []
for i in range(len(a)):
count = 0
for j in range(len(a) - i):
if a[i] != a[j+i]:
out.append(count)
break
else:
count += 1
结果
>>> out
[1, 2, 1, 1, 1, 1, 2, 1, 3, 2, 1]
答案 2 :(得分:0)
如果没有太多独特元素,这里有一个如何对其进行矢量化的想法。它可能不够通用,只能实际解决你的问题序列a
:)(我玩数组直到它工作)。最后的while循环应该是可以优化的东西:
import numpy as np
a = np.array(['a','b','b','c','d','a','b','b','c','c','c','d'])
aa = a[:, np.newaxis] == np.unique(a)
aaa = np.cumsum(aa[::-1], axis=0)[::-1] * aa
# this is where it gets messy
negative_jump = True
while negative_jump:
d = np.diff(aaa, axis=0)
correction = (d + 1) * (d < -1)
negative_jump = (correction != 0).any()
aaa[:-1] += correction
result = aaa[:-1].sum(axis=1)
说明:在循环之前查看aaa
。它将包含远离0的数字。在这个数据视图中,必须从一行传递到另一行,减量永远不会是< -1
。如果是,那么上面的数字太大了。循环将其减小,直到它为-1或0.再次,不是最优的,绝对可以做得更好。