Question

我有一个websocket feed返回列表或列表列表，我想只提取符合我想要的标准的个别列表。

示例数据：

[['o', 0, '9726.85438804', '0.40716427']]
[['o', 0, '9909.02930523', '0.00000000'], ['o', 0, '9855.53483394', '12.07669000']]
[['o', 0, '9669.50657753', '0.00000000'], ['o', 0, '9669.60657753', '0.30000000']]
[['o', 1, '9660.00000000', '0.00000000'], ['o', 1, '9658.52360670', '0.00000000'], ['o', 1, '9652.51635910', '0.00000000'], ['o', 1, '9652.50517986', '0.00244445'], ['t', '23068523', 0, '9660.00000000', '0.00191511', 1525594948], ['t', '23068524', 0, '9660.00000000', '0.00160000', 1525594948], ['t', '23068525', 0, '9658.52360670', '0.00020655', 1525594948], ['t', '23068526', 0, '9652.51635910', '0.00020668', 1525594948], ['t', '23068527', 0, '9652.50517986', '0.10115560', 1525594948]]
[['o', 0, '9727.41662891', '0.58301369']]
[['o', 1, '9642.52790569', '0.00000000'], ['o', 1, '9652.52790569', '0.40000000']]
[['o', 1, '9650.82850048', '0.00000000']]
[['o', 0, '9669.60657753', '0.29000000'], ['t', '23068528', 1, '9669.60657753', '0.01000000', 1525594956]]

我想提取/打印任何具有第一个索引't'的子列表。因此，从上面它将返回：

['t', '23068523', 0, '9660.00000000', '0.00191511', 1525594948]
['t', '23068524', 0, '9660.00000000', '0.00160000', 1525594948]
['t', '23068525', 0, '9658.52360670', '0.00020655', 1525594948]
['t', '23068526', 0, '9652.51635910', '0.00020668', 1525594948]
['t', '23068527', 0, '9652.50517986', '0.10115560', 1525594948]
['t', '23068528', 1, '9669.60657753', '0.01000000', 1525594956]

Answer 1

简单循环

for l in nested_list:
    if l[0] == 't':
        print (l)

如果有外部列表，只需添加一个循环。一个带过滤器的衬管

filter(lambda x: x[0] == 't', nested_list)

大量列表清单 - 仅返回第一个索引匹配的列表

1 个答案: