python3中的Regex：匹配数字或可选句点之后但在可选comm之前的所有内容

Question

我试着从配方中退回原料，没有任何测量或指导。成分列表如下：

['1  medium tomato, cut into 8 wedges',
 '4  c. torn mixed salad greens',
 '1/2  small red onion, sliced and separated into rings',
 '1/4  small cucumber, sliced',
 '1/4  c. sliced pitted ripe olives',
 '2  Tbsp. reduced-calorie Italian salad dressing',
 '2  Tbsp. lemon juice',
 '1  Tbsp. water',
 '1/2  tsp. dried mint, crushed',
 '1/4  c. crumbled Feta cheese or 2 Tbsp. crumbled Blue cheese']

我想返回以下列表：

^{pr2}$

我发现的最接近的模式是：

pattern = '[\s\d\.]* ([^\,]+).*'

但在测试中：

for ing in ingredients:
    print(re.findall(pattern, ing))

每个度量缩写后面的句点也会返回，例如：

['c. torn mixed salad greens']

同时

pattern = '(?<=\. )[^.]*$'

无法捕获没有句点的实例，如果两者都出现，则捕获逗号，即：

[]
['torn mixed salad greens']
[]
[]
['sliced pitted ripe olives']
['reduced-calorie Italian salad dressing']
['lemon juice']
['water']
['dried mint, crushed']
['crumbled Blue cheese']

提前谢谢你！在

Answer 1

您可以使用此模式：

for ing in ingredients:
    print(re.search(r'[a-z][^.,]*(?![^,])(?i)', ing).group())

图案细节：

^{pr2}$