Question

使用：Python 2.4

目前，我有一个嵌套的for循环，迭代2个列表，并根据两个列表中存在的两个元素进行匹配。找到匹配后，它将从r120Final列表中获取元素，并放入一个名为“r120Delta”的新列表：

for r120item in r120Final:
    for spectraItem in spectraFinal:
    if(str(spectraItem[0]) == r120item[2].strip()) and (str(spectraItem[25]) == r120item[10]):
        r120Delta.append(r120item)
        break

问题在于这是如此缓慢而且列表并不那么深。 R120约为64,000行，Spectra约为150,000行。

r120Final列表是一个嵌套数组，它看起来像这样：

r120Final[0] = [['xxx','xxx','12345','xxx','xxx','xxx','xxx','xxx','xxx','xxx','234567']]
...
r120Final[n] = [['xxx','xxx','99999','xxx','xxx','xxx','xxx','xxx','xxx','xxx','678901']]

spectraFinal列表基本相同，是一个嵌套数组，它看起来像这样：

spectraFinal[0] = [['12345','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','234567']]
...
spectraFinal[0] = [['99999','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','xxx','678901']]

最后，“r120Delta”的原因是我可以在r120Final和r120Delta之间执行列表差异，并检索未匹配的r120数据元素。这是我为这个任务定义的函数，再次，慢：

def listDiff( diffList, completeList ):
    returnList = []
        for completeItem in completeList:
            if not completeItem  in diffList:
                returnList.append(completeItem)
    return returnList

基本上，我对Python知识渊博，但绝不是专家。我正在寻找一些专家来告诉我如何加快速度。任何帮助表示赞赏！

Answer 1

spectra_set = set((str(spectraItem[0]), str(spectraItem[25])) for spectraItem in spectraFinal)

returnList = []
for r120item in r120Final:
    if (r120item[2].strip(), r120item[10]) not in spectra_set:
       returnList.append(r120item)

这会添加与returnList不匹配的所有项目。

你可以在一行（如果你真的想要）这样做

returnList = [r120item for r120item in r120Final 
                 if (r120item[2].strip(), r120item[10]) not in 
                     set((str(spectraItem[0]), str(spectraItem[25])) 
                         for spectraItem in spectraFinal)]

如果您需要完整的spectraItem：

spectra_dict = dict(((str(spectraItem[0]), str(spectraItem[25])), spectraItem) for spectraItem in spectraFinal)
returnList = []
for r120item in r120Final:
    key = (r120item[2].strip(), r120item[10])
    if key not in spectra_dict:
        returnList.append(r120item)
    else:
        return_item = some_function_of(r120item, spectra_dict[key])
        returnList.append(return_item)

嵌套for循环搜索2个列表

1 个答案: