我有一个numpy数组。让我们看看下面的例子
a = [255,1,255,255,1,255,255,255,2,2,255,255,255,2,2,3,255,255,255,3]
在上面的数组中，除了值255之外，还会考虑唯一条目。我们希望在每个唯一条目之间填充值。
结果看起来像
[255,1,1,1,1,255,255,255,2,2,2,2,2,2,2,3,3,3,3,3]
可以很容易地完成。寻找pythonic方式。
非常感谢
答案 0 :(得分：1)
我使用itertools模块中的groupby函数。
我还使用了here中的window函数。
from __future__ import print_function
from itertools import tee, izip, groupby
a = [255,1,255,255,1,255,255,255,2,2,255,255,255,2,2,3,255,255,255,3]
def groupby2(iterable):
'''Used to convert to the second iterable element of "groupby" result to list'''
for i in groupby(iterable):
yield (i[0],list(i[1]))
def window(iterable,n):
els = tee(iterable,n)
for i,el in enumerate(els):
for _ in range(i):
next(el, None)
return izip(*els)
def compress(iterable):
it = window(groupby2(iterable),3)
#Creates the iterator which yield the elements in the following manner: (255, [255]), (1, [1]), (255, [255, 255])
for ge in it:
flag = False #Reset the flag
print('\nWindow: {}'.format(ge))
for value in ge[0][1]: #Yield all the values of the first element of the window
print('A: {}'.format(value))
yield value
if ge[1][0]==255 and ge[0][0]==ge[2][0]: #The central element of the window has to be replaced
flag = True #Flag for correct last window processing
for _ in ge[1][1]: #Replacing the central element of the window
print('B: {}'.format(ge[0][0]))
yield ge[0][0]
next(it,None) #Skip 1 element of the 'it' (which will be advanced by 1 element by for-loop, giving 2 net advances).
#Processing the last 2 elements of the last window.
if flag==False: #The central element of the last window hasn't been processed. Proccessing.
for value in ge[1][1]:
print('C: {}'.format(value))
yield value
for value in ge[2][1]: #The last element of the window.
print('D: {}'.format(value))
yield value
print('\nInput: {}'.format(a))
output = list(compress((a)))
print('Proram output: {}'.format(output))
print('Goal output : {}'.format([255,1,1,1,1,255,255,255,2,2,2,2,2,2,2,3,3,3,3,3]))
代码带有调试消息。我会把它们住在这里，因为它们更容易理解它是如何工作的。如果您不需要它们，请删除它们。
输出结果为：
Input: [255, 1, 255, 255, 1, 255, 255, 255, 2, 2, 255, 255, 255, 2, 2, 3, 255, 255, 255, 3]
Window: ((255, [255]), (1, [1]), (255, [255, 255]))
A: 255
Window: ((1, [1]), (255, [255, 255]), (1, [1]))
A: 1
B: 1
B: 1
Window: ((1, [1]), (255, [255, 255, 255]), (2, [2, 2]))
A: 1
Window: ((255, [255, 255, 255]), (2, [2, 2]), (255, [255, 255, 255]))
A: 255
A: 255
A: 255
Window: ((2, [2, 2]), (255, [255, 255, 255]), (2, [2, 2]))
A: 2
A: 2
B: 2
B: 2
B: 2
Window: ((2, [2, 2]), (3, [3]), (255, [255, 255, 255]))
A: 2
A: 2
Window: ((3, [3]), (255, [255, 255, 255]), (3, [3]))
A: 3
B: 3
B: 3
B: 3
D: 3
Proram output: [255, 1, 1, 1, 1, 255, 255, 255, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3]
Goal output : [255, 1, 1, 1, 1, 255, 255, 255, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3]
<强>更新 这是一个重新考虑的版本：
from __future__ import print_function
from itertools import tee, izip, groupby
def groupby2(iterable):
for i in groupby(iterable):
yield (i[0],len(tuple(i[1])))
def window(iterable,n):
els = tee(iterable,n)
for i,el in enumerate(els):
for _ in range(i):
next(el, None)
return izip(*els)
def subs(iterable):
it = window(groupby2(iterable),3)
for left, middle, right in it:
yield [left[0]]*left[1]
if middle[0]==255 and left[0]==right[0]:
yield [left[0]]*middle[1]
next(it,None)
if not(middle[0]==255 and left[0]==right[0]):
yield [middle[0]]*middle[1]
yield [right[0]]*right[1]
def chained(iterable):
for L in subs(iterable):
for el in L:
yield el
a = [255,1,255,255,1,255,255,255,2,2,255,255,255,2,2,3,255,255,255,3]
print('\nInput: {}'.format(a))
output = list(chained((a)))
print('Proram output: {}'.format(output))
print('Goal output : {}'.format([255,1,1,1,1,255,255,255,2,2,2,2,2,2,2,3,3,3,3,3]))
答案 1 :(得分：1)
不知道pythonic在这里意味着什么，但只是我的两分钱，
import numpy as np
a = np.array([255,1,255,255,1,255,255,255,2,2,255,255,255,2,2,3,255,255,255,3])
# find the locations of the unique numbers
b = np.where(a != 255)[0]
# find out what the unique numbers are
u = a[b]
for i,v in zip(b, u):
try:
if (v == vlast): # found a sandwich
if (i != ilast+1): # make sure it has something in between
a[ilast+1: i] = v
else: # make current unique value as the beginning of next sandwich
vlast, ilast = v, i
except NameError:
# initialize the first match
vlast, ilast = v, i
print(a)
它给出了正确的答案：
[255 1 1 1 1 255 255 255 2 2 2 2 2 2 2 3 3 3 3 3]
答案 2 :(得分：0)
基于numpy的简短解决方案：
import numpy
a = numpy.array([255,1,255,255,1,255,255,255,2,2,255,255,255,2,2,3,255,255,255,3])
b = [(i, numpy.argmax(a == i), len(a) - numpy.argmax(a[::-1] == i)) for i in numpy.unique(a[a < 255])]
for i in b:
a[i[1]:i[2]] = i[0]
其中b是由(unique value, start index, end index + 1)组成的元组列表。
答案 3 :(得分：0)
另一个解决方案是在枚举值列表中使用window function包含2个项目和ifilterfalse。
from __future__ import print_function
from itertools import tee, izip, ifilterfalse
def window(iterable,n):
els = tee(iterable,n)
for i,el in enumerate(els):
for _ in range(i):
next(el, None)
return izip(*els)
def replace(iterable,placeholder=255):
it = enumerate(iterable)
def save_last(iterable):
for i in iterable:
yield i
replace.last_index = i[0] #Save the last value
it = save_last(it)
it = ifilterfalse(lambda x: x[1]==placeholder, it)
for i,(left,right) in enumerate(window(it,2)):
if i==0:
for j in range(left[0]):
yield placeholder
yield left[1]
if right[0]>left[0]+1:
if left[1]==right[1]:
for _ in range(right[0]-left[0]-1):
yield left[1]
else:
for _ in range(right[0]-left[0]-1):
yield placeholder
yield right[1]
if right[0]<replace.last_index:
for i in range(replace.last_index-right[0]):
yield placeholder
a = [255,1,255,255,1,255,255,255,2,2,255,255,255,2,2,3,255,255,255,3,255,255]
print('\nInput: {}'.format(a))
output = list(replace(a))
print('Proram output: {}'.format(output))
print('Goal output : {}'.format([255,1,1,1,1,255,255,255,2,2,2,2,2,2,2,3,3,3,3,3,255,255]))
Here我解释它是如何运作的。