Python为每个lin使用rsplit将字符串转换为表

Question

我是Python新手，需要一些关于字符串的帮助，该字符串如下所示：

string='Starters\nSalad with Greens 14.00\nSalad Goat Cheese 12.75\nMains\nPizza 12.75\nPasta 12.75\n'

需要将它转换成一个更像这样的表：

Category   Dish   Price
Starters   Salad with Greens   14.00
Starters   Salad Goat Cheese   12.75
Mains   Pizza   12.75
Mains  Pasta  12.75

实现这一目标的最佳方法是什么？你知道吗

我想申请字符串.rsplit（“”，2），但无法计算如何使其按行执行。不知道如何将标题重复到一个单独的列中。 任何帮助都将不胜感激。你知道吗

提前谢谢！你知道吗

Answer 1

您可以在Python3中使用基于类的解决方案，并使用运算符重载来获得对数据的附加可访问性：

import re
import itertools
class MealPlan:
    def __init__(self, string, headers):
       self.headers = headers
       self.grouped_data = [d for c, d in [(a, list(b)) for a, b in itertools.groupby(string.split('\n'), key=lambda x:x in ['Starters', 'Mains'])]]
       self.final_grouped_data = list(map(lambda x:[x[0][0], x[-1]], [grouped_data[i:i+2] for i in range(0, len(grouped_data), 2)]))
       self.final_data = [[[a, *list(filter(None, re.split('\s(?=\d)', i)))] for i in b] for a, b in final_grouped_data]
       self.final_data = [list(filter(lambda x:len(x) > 1, i)) for i in self.final_data]
    def __getattr__(self, column):
        if column not in self.headers:
            raise KeyError("'{}' not found".format(column))
        transposed = [dict(zip(self.headers, i)) for i in itertools.chain.from_iterable(self.final_data)]
        yield from map(lambda x:x[column], transposed)
    def __getitem__(self, row):
         new_grouped_data = {a:dict(zip(self.headers[1:], zip(*[i[1:] for i in list(b)]))) for a, b in itertools.groupby(list(itertools.chain(*self.final_data)), key=lambda x:x[0])}
         return new_grouped_data[row]
    def __repr__(self):
         return ' '.join(self.headers)+'\n'+'\n'.join('\n'.join(' '.join(c) for c in i) for i in self.final_data)

string='Starters\nSalad with Greens 14.00\nSalad Goat Cheese 12.75\nMains\nPizza 12.75\nPasta 12.75\n' 
meal = MealPlan(string, ['Category', 'Dish', 'Price'])
print(meal)
print([i for i in meal.Category])
print(meal['Starters'])

输出：

Category Dish Price
Starters Salad with Greens 14.00
Starters Salad Goat Cheese 12.75
Mains Pizza 12.75
Mains Pasta 12.75
['Starters', 'Starters', 'Mains', 'Mains']
{'Dish': ('Salad with Greens', 'Salad Goat Cheese'), 'Price': ('14.00', '12.75')}

Answer 2

不是说我会在生产环境中使用它，而是为了学术挑战：

import re

string = """Starters
Salad with Greens 14.00
Salad Goat Cheese 12.75
Mains
Pizza 12.75
Pasta 12.75"""

rx = re.compile(r'^(Starters|Mains)', re.MULTILINE)

result = "\n".join(["{}\t{}".format(category, line)
                for parts in [[part.strip() for part in rx.split(string) if part]]
                for category, dish in zip(parts[0::2], parts[1::2])
                for line in dish.split("\n")])
print(result)

这就产生了

Starters    Salad with Greens 14.00
Starters    Salad Goat Cheese 12.75
Mains   Pizza 12.75
Mains   Pasta 12.75

Answer 3

我想你必须决定如何区分类别和项目。我认为一件商品应该有它的价格。这段代码检查是否存在一个点，但您可能应该使用regexp。你知道吗

s = 'Starters\nSalad with Greens 14.00\nSalad Goat Cheese 12.75\nMains\nPizza 12.75\nPasta 12.75'
items = s.split('\n')
# ['Starters', 'Salad with Greens 14.00', 'Salad Goat Cheese 12.75', 'Mains', 'Pizza 12.75', 'Pasta 12.75']

category = ''
menu = {}
for item in items:
    print(item)
    if '.' in item:
        menu[category].append(item)
    else:
        category = item
        menu[category] = []
print(menu)

# {'Starters': ['Salad with Greens 14.00', 'Salad Goat Cheese 12.75'], 'Mains': ['Pizza 12.75', 'Pasta 12.75']}

UPD：您可以替换

if '.' in item:

与

if re.match(r".*\d.\d\d", item):

它正在搜索以1.11结尾的字符串（如果在category name中有缩写，那么它很有用）