来自其他子列表中的子列表的 python 字符串

Question

此函数应该检查

colors

中子列表中的元素是否存在于

food

中包含的子列表中，并根据

categ_cols

中包含的颜色首字母创建字典

def group_by_category(categs_names, categs_subs, text_subs):
    temp_dict = {name: [] for name in categs_names}

    for categ_index, category_list in enumerate(categs_subs):
        for substring in category_list:
            for sublist in text_subs:
                for text in sublist:
                    if substring in text:
                        temp_dict[categs_names[categ_index]].append(text)
                    else:
                        temp_dict[categs_names[categ_index]].append('None')
    return temp_dict

categs_cols = ['y', 'r', 'g']
food = [['banana is yellow', 'apple is red', 'pear is green' ] , ['lettuce is green' ,'pasta is yellowish','sugar is brown'] ]
colors = [[ 'yellow', 'yellowish'  ], ['red'], ['green']]
grouped_cols = group_by_category(categs_cols, colors, food)
print(grouped_cols)

预期结果如下

{'y': ['banana is yellow', 'pasta is yellowish'], 'r': ['None', 'apple is red'], 
 'g': ['pear is green', 'lettuce is green']}

但我却得到了这个

 {'y': ['banana is yellow', 'None', 'None', 'None', 'pasta is yellowish', 'None', 'None', 'None', 'None', 'None', 'pasta is yellowish', 'None'], 'r': ['None', 'apple is red', 'None', 'None', 'None', 'None'], 'g': ['None', 'None', 'pear is green', 'lettuce is green', 'None', 'None']}

我尝试在下面使用

break

if substring in text:   
    temp_dict[categs_names[categ_index]].append(text)

但是没有效果

Answer 1

每当食物项与当前关键字不匹配时，else 语句将添加 None 。相反，您想要的是检查当前category_list的最末尾（即最外层循环的末尾）是否匹配列表为空，然后添加 None 。结果应该是这样的：

def group_by_category(categs_names, categs_subs, text_subs):
    temp_dict = dict()

    for categ_index, category_list in enumerate(categs_subs):
        matches = [] # Here is the list of matches
        for substring in category_list:
            for sublist in text_subs:
                for text in sublist:
                    if substring in text:
                        matches.append(text)
        if len(matches) == 0: # Only add None if the search came up empty
            matches.append("None")
        temp_dict[categs_names[categ_index]] = matches
    return temp_dict

categs_cols = ['y', 'r', 'g', 'e']
food = [['banana is yellow', 'apple is red', 'pear is green' ] , ['lettuce is green' ,'pasta is yellowish','sugar is brown']]
colors = [[ 'yellow', 'yellowish'  ], ['red'], ['green'], ['empty']]
grouped_cols = group_by_category(categs_cols, colors, food)
print(grouped_cols)

""" >>> {'y': ['banana is yellow', 'pasta is yellowish', 'pasta is yellowish'], 'r': ['apple is red'], 'g': ['pear is green', 'lettuce is green'], 'e': ['None']} """

Answer 2

如果您通过

zip()

和扁平化理解进行一些输入重塑，那么您可以执行以下操作：

def group_by_category(categs_names, categs_subs, text_subs):
    retval = {}
    for categs_id, categs_words in zip(categs_names, categs_subs):
        for text_sub in [cell for row in text_subs for cell in row]:
            if any(word in text_sub for word in categs_words):
                retval.setdefault(categs_id, []).append(text_sub)
    return retval

categs_cols = [
    'y',
    'r',
    'g'
]

colors = [
    ['yellow', 'yellowish'],
    ['red'],
    ['green']
]

food = [
    ['banana is yellow', 'apple is red', 'pear is green' ],
    ['lettuce is green' ,'pasta is yellowish','sugar is brown']
]

print(group_by_category(categs_cols, colors, food))

给你：

{
    'y': ['banana is yellow', 'pasta is yellowish'],
    'r': ['apple is red'],
    'g': ['pear is green', 'lettuce is green']
}