在python中将文件解析为字典

I have a file, a small fragment of it you can see below:

我有一个文件，你可以看到它的一小部分：

Clutch001
Albino X Pastel
Bumble Bee X Albino Lesser
Clutch002
Bee X Fire Bee
Albino Cinnamon X Albino
Mojave X Bumble Bee
Clutch003
Black Pastel X Banana Ghost Lesser
....

Number of strings between ClucthXXX and next ClutchXXX might be different but not equal to zero. I was wondering if it's possible somehow to take a specific string from a file using it as a key (in my case it would be ClutchXXX) and the text till the second occurrence of the specific string as a value for a dictionary? I want to receive such dictionary:

ClucthXXX和下一个ClutchXXX之间的字符串数可能不同但不等于零。我想知道是否有可能以某种方式从文件中使用它作为键（在我的情况下它将是ClutchXXX）和文本，直到第二次出现特定字符串作为字典的值？我想收到这样的字典：

d={'Clutch001': 'Albino X Pastel, Bumble Bee X Albino Lesser'
   'Clutch002': 'Bee X Fire Bee, Albino Cinnamon X Albino, Mojave X Bumble Bee'
   'Clutch003': 'Black Pastel X Banana Ghost Lesser'}

I am mostly interested in the part where we take string pattern and save it as a key and the text after as a value. Any suggestions or directions to a useful approach would be appreciated.

我最感兴趣的是我们采用字符串模式的部分，并将其保存为键，将文本作为值保存。任何有用的方法的建议或指示将不胜感激。

7 个解决方案

#1

from itertools import groupby
from functools import partial

key = partial(re.match, r'Clutch\d\d\d')

with open('foo.txt') as f:
    groups = (', '.join(map(str.strip, g)) for k, g in groupby(f, key=key))
    pprint(dict(zip(*[iter(groups)]*2)))

{'Clutch001': 'Albino X Pastel, Bumble Bee X Albino Lesser',
 'Clutch002': 'Bee X Fire Bee, Albino Cinnamon X Albino, Mojave X Bumble Bee',
 'Clutch003': 'Black Pastel X Banana Ghost Lesser'}

#2

Collect the lines in lists, storing that list in a dictionary at the same time:

收集列表中的行，同时将该列表存储在字典中：

d = {}
values = None
with open(filename) as inputfile:
    for line in inputfile:
        line = line.strip()
        if line.startswith('Clutch'):
            values = d[line] = []
        else:
            values.append(line)

This gives you:

这给你：

{'Clutch001': ['Albino X Pastel', 'Bumble Bee X Albino Lesser']
 'Clutch002': ['Bee X Fire Bee', 'Albino Cinnamon X Albino', 'Mojave X Bumble Bee']
 'Clutch003': ['Black Pastel X Banana Ghost Lesser']}

It's easy enough to turn all those lists into single strings though, after loading the file:

在加载文件后，将所有这些列表转换为单个字符串很容易：

d = {key: ', '.join(value) for key, value in d.items()}

You can also do the joining as you read the file; I'd use a generator function to process the file in groups:

您也可以在阅读文件时进行加入;我将使用生成器函数来处理组中的文件：

def per_clutch(inputfile):
    clutch = None
    lines = []
    for line in inputfile:
        line = line.strip()
        if line.startswith('Clutch'):
            if lines:
                yield clutch, lines
            clutch, lines = line, []
        else:
            lines.append(line)
    if clutch and lines:
        yield clutch, lines

then just slurp all groups into a dictionary:

然后只是将所有组塞进字典：

with open(filename) as inputfile:
    d = {clutch: ', '.join(lines) for clutch, lines in per_clutch(inputfile)}

Demo of the latter:

演示后者：

>>> def per_clutch(inputfile):
...     clutch = None
...     lines = []
...     for line in inputfile:
...         line = line.strip()
...         if line.startswith('Clutch'):
...             if lines:
...                 yield clutch, lines
...             clutch, lines = line, []
...         else:
...             lines.append(line)
...     if clutch and lines:
...         yield clutch, lines
... 
>>> sample = '''\
... Clutch001
... Albino X Pastel
... Bumble Bee X Albino Lesser
... Clutch002
... Bee X Fire Bee
... Albino Cinnamon X Albino
... Mojave X Bumble Bee
... Clutch003
... Black Pastel X Banana Ghost Lesser
... '''.splitlines(True)
>>> {clutch: ', '.join(lines) for clutch, lines in per_clutch(sample)}
{'Clutch001': 'Albino X Pastel, Bumble Bee X Albino Lesser', 'Clutch002': 'Bee X Fire Bee, Albino Cinnamon X Albino, Mojave X Bumble Bee', 'Clutch003': 'Black Pastel X Banana Ghost Lesser'}
>>> from pprint import pprint
>>> pprint(_)
{'Clutch001': 'Albino X Pastel, Bumble Bee X Albino Lesser',
 'Clutch002': 'Bee X Fire Bee, Albino Cinnamon X Albino, Mojave X Bumble Bee',
 'Clutch003': 'Black Pastel X Banana Ghost Lesser'}

#3

As noted in comments, if "Clutch" (or whatever keyword) can be relied on not to appear in the non-keyword lines, you could use the following:

如评论中所述，如果可以依赖“离合器”（或任何关键字）不出现在非关键字行中，则可以使用以下内容：

keyword = "Clutch"
with open(filename) as inputfile:
    t = inputfile.read()
    d = {keyword + s[:3]: s[3:].strip().replace('\n', ', ') for s in t.split(keyword)}

This reads the whole file in to memory at once, so should be avoided if your file may get very large.

这会立即将整个文件读入内存，因此如果您的文件可能变得非常大，应该避免使用。

#4

You could use re.split() to enumerate "Clutch" parts in the file:

您可以使用re.split（）枚举文件中的“Clutch”部分：

import re

tokens = iter(re.split(r'(^Clutch\d{3}\s*$)\s+', file.read(), flags=re.M))
next(tokens) # skip until the first Clutch
print({k: ', '.join(v.splitlines()) for k, v in zip(tokens, tokens)})

Output

{'Clutch001': 'Albino X Pastel, Bumble Bee X Albino Lesser', 
 'Clutch002': 'Bee X Fire Bee, Albino Cinnamon X Albino, Mojave X Bumble Bee',
 'Clutch003': 'Black Pastel X Banana Ghost Lesser'}

#5

Lets file 'file.txt' contains:

让文件'file.txt'包含：

Clutch001
Albino X Pastel
Bumble Bee X Albino Lesser
Clutch002
Bee X Fire Bee
Albino Cinnamon X Albino
Mojave X Bumble Bee
Clutch003
Black Pastel X Banana Ghost Lesser

To receive your dictionary try this:

要收到你的字典，试试这个：

import re

with open('file.txt', 'r') as f:
    result = re.split(
        r'(Clutch\d{3}).*?',
        f.read(),
        flags=re.DOTALL # including '\n'
    )[1:] # result is ['Clutch001', '\nAlbino X Pastel\nBumble Bee X Albino Lesser\n', 'Clutch002', '\nBee X Fire Bee\nAlbino Cinnamon X Albino\nMojave X Bumble Bee\n', 'Clutch003', '\nBlack Pastel X Banana Ghost Lesser\n']

    keys = result[::2] # keys is ['Clutch001', 'Clutch002', 'Clutch003']
    values = result[1::2] # values is ['\nAlbino X Pastel\nBumble Bee X Albino Lesser\n', '\nBee X Fire Bee\nAlbino Cinnamon X Albino\nMojave X Bumble Bee\n', '\nBlack Pastel X Banana Ghost Lesser\n']

    values = map(
        lambda value: value.strip().replace('\n', ', '),
        values
    ) # values is ['Albino X Pastel, Bumble Bee X Albino Lesser', 'Bee X Fire Bee, Albino Cinnamon X Albino, Mojave X Bumble Bee', 'Black Pastel X Banana Ghost Lesser']

    d = dict(zip(keys, values)) # d is {'Clutch001': 'Albino X Pastel, Bumble Bee X Albino Lesser', 'Clutch002': 'Bee X Fire Bee, Albino Cinnamon X Albino, Mojave X Bumble Bee', 'Clutch003': 'Black Pastel X Banana Ghost Lesser'}

#6

Here's a version that works, more or less. I'm not sure how Pythonic it is (it can probably be squeezed and can definitely be improved):

这是一个或多或少有效的版本。我不确定它是如何Pythonic（它可能被挤压，绝对可以改进）：

import re
import fileinput

d = dict()
key = ''
rx = re.compile('^Clutch\d\d\d$')

for line in fileinput.input():
    line = line[0:-1]
    if rx.match(line):
        key = line
        d[key] = ''
    else:
        d[key] += line

print d

for key in d:
    print key, d[key]

The output (which repeats the information) is:

输出（重复信息）是：

{'Clutch001': 'Albino X PastelBumble Bee X Albino Lesser', 'Clutch002': 'Bee X Fire BeeAlbino Cinnamon X AlbinoMojave X Bumble Bee', 'Clutch003': 'Black Pastel X Banana Ghost Lesser'}
Clutch001 Albino X PastelBumble Bee X Albino Lesser
Clutch002 Bee X Fire BeeAlbino Cinnamon X AlbinoMojave X Bumble Bee
Clutch003 Black Pastel X Banana Ghost Lesser

If for some reason the first line isn't a 'clutch' line, you get an error because of the empty key.

如果由于某种原因第一行不是“离合器”行，则由于空键而出现错误。

Joining with commas, dealing with broken text files (no newline at the end) etc:

加入逗号，处理损坏的文本文件（最后没有换行符）等：

import fileinput

d = {}

for line in fileinput.input():
    line = line.rstrip('\r\n') # line.strip() for leading and trailing space
    if line.startswith('Clutch'):
        key = line
        d[key] = ''
        pad = ''
    else:
        d[key] += pad + line
        pad = ', '

print d

for key in d:
    print "'%s': '%s'" % (key, d[key])

The 'pad' technique is one I like in other contexts, and it works fine here. I'm tolerably certain it wouldn't be regarded as Pythonic, though.

'pad'技术是我在其他环境中喜欢的技术，它在这里工作得很好。不过，我可以肯定它不会被视为Pythonic。

Revised sample output:

修改样本输出：

{'Clutch001': 'Albino X Pastel, Bumble Bee X Albino Lesser', 'Clutch002': 'Bee X Fire Bee, Albino Cinnamon X Albino, Mojave X Bumble Bee', 'Clutch003': 'Black Pastel X Banana Ghost Lesser'}
'Clutch001': 'Albino X Pastel, Bumble Bee X Albino Lesser'
'Clutch002': 'Bee X Fire Bee, Albino Cinnamon X Albino, Mojave X Bumble Bee'
'Clutch003': 'Black Pastel X Banana Ghost Lesser'

#7

Assuming the word Clutch occurs independently on its own line, the following will work:

假设Clutch这个词在它自己的行上独立发生，以下内容将起作用：

import re
d = {}
with open(filename) as f:
for line in f:
    if re.match("^Clutch[0-9]+", line) :
        match = line   # match is the key searched for
        match = match.replace('\n', ' ')    # newlines are replaced
        d[match] = ''
    else:
        line = line.replace('\n', ' ')
        d[match] += line  # all lines without the word 'Clutch'
                          # are added to the matched key