如何从分隔列表创建关联数组?

2024-05-15 05:42:32 发布

您现在位置:Python中文网/ 问答频道 /正文

我正在处理从存储设备读出的数据,格式如下:

id:name:UPS_serial_number:WWNN:status:IO_group_id:IO_group_name:config_node:UPS_unique_id:hardware:iscsi_name:iscsi_alias:panel_name:enclosure_id:canister_id:enclosure_serial_number:site_id:site_name
10:node_A::00A550:online:0:io_grp0:yes::SV1:iqn.1986-03.com:2145.test.nodeA::A:::::
15:node_B::00A548:online:0:io_grp0:no::SV1:iqn.1986-03.com.:2145.test.nodeB::B:::::

如何将数据作为2D数组读取,比如datarray['15']['status']?在

我试过这样做:

^{pr2}$

但这似乎把数据分割错了:

[[[0, u'id'], [1, u'name'], [2, u'UPS_serial_number'], [3, u'WWNN'], [4, u'status'], [5, u'IO_group_id'], [6, u'IO_group_name'], [7, u'config_node'], [8, u'UPS_unique_id'], [9, u'hardware'], [10, u'iscsi_name'], [11, u'iscsi_alias'], [12, u'panel_name'], [13, u'enclosure_id'],

Tags: 数据nameioidconfignodenumberstatus
2条回答

使用^{}将各行作为字典读取,然后使用字典理解创建“外部”dict,将ID属性映射到具有该ID的内部dict

raw = """id:name:UPS_serial_number:WWNN:status:IO_group_id:IO_group_name:config_node:UPS_unique_id:hardware:iscsi_name:iscsi_alias:panel_name:enclosure_id:canister_id:enclosure_serial_number:site_id:site_name
10:node_A::00A550:online:0:io_grp0:yes::SV1:iqn.1986-03.com:2145.test.nodeA::A:::::
15:node_B::00A548:online:0:io_grp0:no::SV1:iqn.1986-03.com.:2145.test.nodeB::B:::::"""

reader = csv.DictReader(raw.splitlines(), delimiter=":")
result = {line["id"]: line for line in reader}
print(result["15"]["status"])  # 'online'

请注意,这不是二维数组,而是字典字典(字典是关联数组)。作为一个简单的2D数组,像result["15"]["status"]这样的查询将不起作用。在

我可以从数据中看出,它是冒号(:)分隔的数据,第一行有头。如果是这种情况,您可以在加载带有分隔符=':'的csv文件时将其加载到pandas dataframe。然后将该数据帧转换为numpy数组。在

import pandas as pd
import os
os.chdir('/Users/Downloads/')
df = pd.read_csv('train.txt',sep=':')
df

id  name    UPS_serial_number   WWNN    status  IO_group_id IO_group_name   config_node UPS_unique_id   hardware    iscsi_name  iscsi_alias panel_name  enclosure_id    canister_id enclosure_serial_number site_id site_name
10  node_A  NaN 00A550  online  0   io_grp0 yes NaN SV1 iqn.1986-03.com 2145.test.nodeA NaN A   NaN NaN NaN NaN NaN
15  node_B  NaN 00A548  online  0   io_grp0 no  NaN SV1 iqn.1986-03.com.    2145.test.nodeB NaN B   NaN NaN NaN NaN NaN

df.as_matrix()

array([['node_A', nan, '00A550', 'online', 0, 'io_grp0', 'yes', nan,
        'SV1', 'iqn.1986-03.com', '2145.test.nodeA', nan, 'A', nan, nan,
        nan, nan, nan],
       ['node_B', nan, '00A548', 'online', 0, 'io_grp0', 'no', nan,
        'SV1', 'iqn.1986-03.com.', '2145.test.nodeB', nan, 'B', nan, nan,
        nan, nan, nan]], dtype=object)

相关问题 更多 >

    热门问题