根据预定义的字符类型拆分字符串

Question

我有一个预定义的字符->类型字典。例如，'a' - 是小写字母，1 是数字，')' 是标点符号等。 使用以下脚本，我标记给定字符串中的所有字符：

labels=''
for ch in list(example):
    try:
        l = character_type_dict[ch]
        print(l)
        labels = labels+l
    except KeyError:
        labels = labels+'o'
        print('o')
labels

例如，给定"1,234.45kg (in metric system)" 作为输入，代码将生成dpdddpddwllwpllwllllllwllllllp 作为输出。

现在，我想根据组拆分字符串。输出应该是这样的：

['1',',','234','.','45','kg',' ','(','in',' ','metric',' ','system',')']

也就是说，它应该根据字符类型的边框进行分割。 有什么想法可以有效地做到这一点吗？

Answer 1

labels 是错误的（在您的示例中是 'dpdddpddwllwpllwllllllwllllllp'，但我认为应该是 'dpdddpddllwpllwllllllwllllllp'）

无论如何，你可以使用滥用itertools.groupby：

from itertools import groupby

example = "1,234.45kg (in metric system)"
labels = 'dpdddpddllwpllwllllllwllllllp'

output = [''.join(group)
          for _, group in groupby(example, key=lambda ch: labels[example.index(ch)])]

print(output)
# ['1', ',', '234', '.', '45', 'kg', ' ', '(', 'in', ' ', 'metric', ' ', 'system', ')']

Answer 2

您可以更简洁地计算标签（而且很可能更快）：

labels = ''.join(character_type_dict.get(ch, 'o') for ch in example)

或者，使用辅助函数：

character_type = lambda ch: character_type_dict.get(ch, 'o')
labels = ''.join(map(character_type, example))

但是你不需要标签来分割字符串；在 itertools.groupby 的帮助下，你可以直接拆分：

splits = list(''.join(g)
              for _, g in itertools.groupby(example, key=character_type)

一个可能更有趣的结果是类型元组和相关分组的向量：

 >>> list((''.join(g), code)
 ...      for code, g in itertools.groupby(example, key=character_type))
 [('1', 'd'), (',', 'p'), ('234', 'd'), ('.', 'p'), ('45', 'd'), ('kg', 'l'),
  (' ', 'w'), ('(', 'p'), ('in', 'l'), (' ', 'w'), ('metric', 'l'), (' ', 'w'),
  ('system', 'l'), (')', 'p')]

---

我计算character_type_dict如下：

character_type_dict = {}
for code, chars in (('w', string.whitespace),
                    ('l', string.ascii_letters),
                    ('d', string.digits),
                    ('p', string.punctuation)):
  for char in chars: character_type_dict[char] = code

但我也可以这样做（我后来发现）：

from collections import ChainMap
character_type_dict = dict(ChainMap(*({c:t for c in string.__getattribute__(n)}
                                    for t,n in (('w', 'whitespace')
                                               ,('d', 'digits')
                                               ,('l', 'ascii_letters')
                                               ,('p', 'punctuation')))))

Answer 3

只记得最后一个类型的类：

import string
character_type = {c: "l" for c in string.ascii_letters}
character_type.update({c: "d" for c in string.digits})
character_type.update({c: "p" for c in string.punctuation})
character_type.update({c: "w" for c in string.whitespace})

example = "1,234.45kg (in metric system)"

x = []
prev = None
for ch in example:
    try:
        l = character_type[ch]
        if l == prev:
            x[-1].append(ch)
        else:
            x.append([ch])
    except KeyError:
        print(ch)
    else:
        prev = l
x = map(''.join, x)
print(list(x))
# ['1', ',', '234', '.', '45', 'kg', ' ', '(', 'in', ' ', 'metric', ' ', 'system', ')']

Answer 4

另一种算法方法。使用 dictionaryget(value, default_value) 方法而不是 try: except: 更好。

import string

character_type_dict = {}
for ch in string.ascii_lowercase:
    character_type_dict[ch] = 'l'
for ch in string.digits:
    character_type_dict[ch] = 'd'
for ch in string.punctuation:
    character_type_dict[ch] = 'p'
for ch in string.whitespace:
    character_type_dict[ch] = 'w'

example = "1,234.45kg (in metric system)"

split_list = []
split_start = 0
for i in range(len(example) - 1):
    if character_type_dict.get(example[i], 'o') != character_type_dict.get(example[i + 1], 'o'):
        split_list.append(example[split_start: i + 1])
        split_start = i + 1
split_list.append(example[split_start:])

print(split_list)

Answer 5

将此作为算法难题：

# dummy mapping
character_type_dict = dict({c: "l" for c in string.ascii_letters}.items()  \
                         + {c: "d" for c in string.digits}.items() \
                         + {c: "p" for c in string.punctuation}.items() \
                         + {c: "w" for c in string.whitespace}.items())
example = "1,234.45kg (in metric system)"
last = example[0]
temp = last
res = []
for ch in example[1:]:
  try:
    cur = character_type_dict[ch]
    if cur != last:
      res.append(temp)
      temp = ''
    temp += ch
    last = cur
  except KeyError:
    last = 'o'
res.append(temp)

结果：

['1', ',', '234', '.', '45', 'kg', ' ', '(', 'in', ' ', 'metric', ' ', 'system', ')']