smerèu prods是一个字典,它看起来像这样:
smer_prods = {
#'vermicelli' : ['vermicelli', 'ragi vermicelli', 'rice vermicelli'],
'ragi vermicelli' : ['ragi vermicelli'],
'rice vermicelli' : ['rice vermicelli'],
'vermicelli jupiter' : ['vermicelli jupiter'],
'lemon & tamarind vermicelli' : ['lemon & tamarind vermicelli'],
'finosta vermicelli' : ['finosta vermicelli-5kg'],
'rosted vermicelli' : ['roasted vermicelli'],
'semiya/vermicelli' : ['semiya / vermicelli 900grams'],
'vermicelli upma' : ['vermicelli upma'],
'vermicelli payasam mix' : ['vermicelli payasam mix'],
'mung bean vermicelli' : ['mung bean vermicelli'],
'red chili' : ['red chilli (lal mirch)','guntur red chilli','red chilly whole(lal mirch)', 'red chilly wg', 'red chilli whole (hot) 1 kg', 'red chilli whole (rich colour) 1 kg'],
'red chili powder' : ['red chilli fresh-kg','red chilli powder (rich colour) 1 kg','red chilli powder (hot) 1 kg','red chilli powder','lal mirch powder','lal mirch powder 100gms', 'lal mirch powder 1kg', 'lal mirch powder 200gms', 'lal mirch powder 500gms'],
'red chilli sauce' : ['red chilli sauce', 'red chilli sauce 200gm pet bottle 48X200gm', 'hot chili sauce'],
'sriraja hot chilli sauce' : ['sriraja hot chilli sauce', 'sriracha hot chilli sauce'],
'mineral water' : ['himalayan orchard pure peach flavoured natural mineral water - 500 ml','himalayan orchard pure strawberry flavoured natural mineral water - 500 ml','himalayan orchard pure apple flavoured natural mineral water - 500 ml','himalayan - the natural mineral water - 500ml bottle', 'himalayan - the natural mineral water - 200ml bottle', 'himalayan - the natural mineral water - 1ltr bottle'],
}
现在当我执行这个简单的代码时:
for x in smer_prods:
mask = df['ITEM NAME'] == x
df1 = df[mask]
print(df1['ITEM NAME'])
它应该只打印ITEM\u NAME列,但它会打印很多不必要的额外信息:
4 rice vermicelli
Name: ITEM NAME, dtype: object
Series([], Name: ITEM NAME, dtype: object)
Series([], Name: ITEM NAME, dtype: object)
5 ragi vermicelli
Name: ITEM NAME, dtype: object
Series([], Name: ITEM NAME, dtype: object)
Series([], Name: ITEM NAME, dtype: object)
Series([], Name: ITEM NAME, dtype: object)
Series([], Name: ITEM NAME, dtype: object)
Series([], Name: ITEM NAME, dtype: object)
Series([], Name: ITEM NAME, dtype: object)
Series([], Name: ITEM NAME, dtype: object)
Series([], Name: ITEM NAME, dtype: object)
1 sriraja hot chilli sauce
Name: ITEM NAME, dtype: object
Series([], Name: ITEM NAME, dtype: object)
Series([], Name: ITEM NAME, dtype: object)
我只想要列表中匹配的项目。为此我写了一段代码:
for x in smer_prods:
mask = df['ITEM NAME'] == x
df1 = df[mask]
item_name = df1['ITEM NAME'].tolist()
print(item_name)
它产生了这样的结果: [“米粉”]
[]
[]
['ragi vermicelli']
[]
[]
[]
[]
[]
[]
[]
[]
['sriraja hot chilli sauce']
[]
[]
期望输出:
['rice vermicelli', 'ragi vermicelli', 'sriraja hot chili sauce']
编辑: 如果我写这个:
match = []
for x in smer_prods:
mask = df['ITEM NAME'] == x
if mask.any() == True:
df1 = df[mask]
item_name = df1['ITEM NAME']#note
match.append(item_name)
print(match)
print('-'*80)
它输出:
[4 rice vermicelli
Name: ITEM NAME, dtype: object, 5 ragi vermicelli
Name: ITEM NAME, dtype: object, 1 sriraja hot chilli sauce
Name: ITEM NAME, dtype: object]
我认为您需要列表中的测试值,所以使用列表理解和
isin
作为字典的测试值:回路解决方案:
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