Question

我正在使用python 2.7，并使用python dict s。

我有这样的输出：

goods: apples, oranges
trunk_names: trunk01, trunk02, trunk03,trunk04,
             trunk05,trunk06, trunk07,trunk08,
             trunk09,trunk10, trunk11,trunk12

我的代码：

    d = {}
    for line in output.split("\n"):
        if ":" not in line:
            continue
        key, value = line.strip().split(":", 1)
        d[key] = value

预期密钥及其值：

 trunk_names: trunk01, trunk02, trunk03,trunk04,trunk05,trunk06, trunk07,trunk08,trunk09,trunk10, trunk11,trunk12

输出实际的键和值：

 trunk_names: trunk01, trunk02, trunk03,trunk04,

Answer 1

from collections import defaultdict

output = '''
goods: apples, oranges
trunk_names: trunk01, trunk02, trunk03,trunk04,
             trunk05,trunk06, trunk07,trunk08,
             trunk09,trunk10, trunk11,trunk12
'''

d = defaultdict(list)

current_key = None

for line in output.split('\n')[1:]:
    if ":" in line:
        current_key = line.split(':')[0].strip()
        values = line.split(':')[1]
    else:
        values = line

    d[current_key] += [
        value.strip()
        for value in values.split(',')
        if value.strip()
    ]


print(d)

给出：

defaultdict(<type 'list'>, {'trunk_names': ['trunk01', 'trunk02', 'trunk03', 'trunk04', 'trunk05', 'trunk06', 'trunk07', 'trunk08', 'trunk09', 'trunk10', 'trunk11', 'trunk12'], 'goods': ['apples', 'oranges']})

Answer 2

在你现有的代码中，你正在做的是跳过没有冒号的行，所以你丢弃了一半以上的数据。

您可以使用<key : list>对词典将每个项目存储在由该键索引的列表中。

from collections import defaultdict
import pprint

output = '''goods: apples, oranges
trunk_names: trunk01, trunk02, trunk03,trunk04,
             trunk05,trunk06, trunk07,trunk08,
             trunk09,trunk10, trunk11,trunk12'''

d = defaultdict(list)
for line in output.split("\n"):
    if ":" in line:
        key, line = line.split(':')

    items = line.strip().split(',')
    d[key].extend([i.strip() for i in items if i]) # check to make sure we don't add '' items to the list

pprint.pprint(d)

在这里，我使用collections.defaultdict，因此您无需在向其添加项目之前显式创建密钥条目。

运行此命令：

{'goods': ['apples', ' oranges'],
 'trunk_names': ['trunk01',
                 'trunk02',
                 'trunk03',
                 'trunk04',
                 'trunk05',
                 'trunk06',
                 'trunk07',
                 'trunk08',
                 'trunk09',
                 'trunk10',
                 'trunk11',
                 'trunk12']}

Answer 3

您的结构有多稳定，如果它非常稳定且数据质量很高，那么您可以通过测试line.endswith(',')：

进行简化

In []:
d = {}
f = iter(output.split('\n'))
for line in f:
    key, line = map(str.strip, line.split(':', 1))
    while line.endswith(','):
        line += next(f)
    d[key] = [i.strip() for i in line.split(',')]

pprint.pprint(d)

Out[]:
{'goods': ['apples', 'oranges'],
 'trunk_names': ['trunk01',
                 'trunk02',
                 'trunk03',
                 'trunk04',
                 'trunk05',
                 'trunk06',
                 'trunk07',
                 'trunk08',
                 'trunk09',
                 'trunk10',
                 'trunk11',
                 'trunk12']}

在单独的行中将逗号分隔的项添加到Python中的dict

3 个答案: