如何在两个大列表之间进行比较快速python
如何在两个大列表之间进行比较快速python
提问于 2014-05-19 20:58:04
我想知道是否有任何方法使这段代码运行得更快。它花了我47秒，它必须比较所有的东西，而不仅仅是在同一位置的元素。
pixels = list(mensagem)
arrayBits = []
for i in pixels:
for j in tabela:
if i == j[0]:
arrayBits.append(j[1])这是漏洞代码，但我认为唯一需要这么长时间的原因是我问的那个。对不起我的英语，我是葡萄牙人。
def codifica(mensagem, tabela, filename):
tamanho = np.shape(mensagem)
largura = tamanho[0]
if len(tamanho)==2:
altura = tamanho[1]
else:
altura = 0
pixels = list(mensagem)
arrayBits = []
for i in pixels:
for j in tabela:
if i == j[0]:
arrayBits.append(j[1])
arraySemVirgulas = np.array(arrayBits).ravel() # tirar as virgulas
arrayJunto = ''.join(arraySemVirgulas) # juntar todos os bits
array = list(map(int,arrayJunto)) # coloca-los numa lista
count = 0
while(len(array)%8!=0):
array.append(0)
count += 1
array = np.array(array)
arrayNovo = array.reshape(-1,8)
decimais = convBi(arrayNovo)
array_char = ['' for i in range(len(decimais)+5)]
j = 2
for i in decimais:
a = chr(i)
array_char[j] = a
j += 1
array_char[0] = str(count)
array_char[1] = str(len(str(largura)))
array_char[2] = str(len(str(altura)))
array_char[3] = str(largura)
array_char[4] = str(altura)
ficheiro = open(filename,"wb")
for i in array_char:
ficheiro.write(i)
ficheiro.close()回答 2
Stack Overflow用户
回答已采纳
发布于 2014-05-19 21:06:42
如果替换迭代，这可能会更快。
for i in pixels:
for j in tabela:
if i == j[0]:
arrayBits.append(j[1])用字典查找
tabela_dict = dict(tabela)
for i in pixels:
if i in tabela_dict :
arrayBits.append(tabela_dict[i])Stack Overflow用户
发布于 2014-05-19 21:43:51
使用set()和基于dict()的容器可以使其在时间上线性化，而不是O(n^2)。这应该会加快速度：
编辑更简单、可能更快的版本：
import itertools
# set of 'keys' that exist in both
keys = set(pixels) & set(el[0] for el in tabela)
# and generator comprehension with fast lookup
elements = (element[1] for element in tabela
if element[0] in keys)
# this will flatten inner lists and create a list with result:
result = list(itertools.chain.from_iterable(elements))只有两次通过tabela，都具有时间复杂度O(n)。
如果pixels不是唯一的，并且每个像素的出现都应该乘以tabela的相应值，则应该使用以下方法：
import itertools
# set of 'keys' that exist in both
keys = set(pixels) & set(el[0] for el in tabela)
# and generator comprehension with fast lookup
elements = lambda key: tabela[key][1] if key in keys else []
# this will flatten inner lists and create a list with result:
flatten = itertools.chain.from_iterable
result = list(flatten(elements(pixel) for pixel in pixels))页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/23747098
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