我试图优化我的代码,因为当我试图加载巨大的字典时,它变得非常慢。我想是因为它在字典里找一个键。我一直在读pythondefaultdict
,我认为这可能是一个很好的改进,但我没有在这里实现它。如您所见,这是一个层次字典结构。任何暗示都将不胜感激。你知道吗
class Species:
'''This structure contains all the information needed for all genes.
One specie have several genes, one gene several proteins'''
def __init__(self, name):
self.name = name #name of the GENE
self.genes = {}
def addProtein(self, gene, protname, len):
#Converting a line from the input file into a protein and/or an exon
if gene in self.genes:
#Gene in the structure
self.genes[gene].proteins[protname] = Protein(protname, len)
self.genes[gene].updateProts()
else:
self.genes[gene] = Gene(gene)
self.updateNgenes()
self.genes[gene].proteins[protname] = Protein(protname, len)
self.genes[gene].updateProts()
def updateNgenes(self):
#Updating the number of genes
self.ngenes = len(self.genes.keys())
基因和蛋白质的定义如下:
class Protein:
#The class protein contains information about the length of the protein and a list with it's exons (with it's own attributes)
def __init__(self, name, len):
self.name = name
self.len = len
class Gene:
#The class gene contains information about the gene and a dict with it's proteins (with it's own attributes)
def __init__(self, name):
self.name = name
self.proteins = {}
self.updateProts()
def updateProts(self):
#Update number of proteins
self.nproteins = len(self.proteins)
不能使用
defaultdict
,因为__init__
方法需要参数。你知道吗这可能是您的瓶颈之一:
len(self.genes.keys())
在计算长度之前创建所有键的list
。这意味着,每次你添加一个基因,你就创建一个列表,然后扔掉它。你拥有的基因越多,这个列表就越昂贵。要避免创建中间列表,只需执行len(self.genes)
。你知道吗更好的办法是将
ngenes
设为property,这样它只在需要时才计算出来。你知道吗对于
Gene
类中的nproteins
也可以这样做。你知道吗以下是重构后的代码:
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