Question

我有一个看起来像这样的对象列表：

Bidder - Timestamp: 11, User ID: 8, Action: BID, Loan ID: 430, Rate: 0.20 -- missed
Bidder - Timestamp: 13, User ID: 13, Action: BID, Loan ID: 430, Rate: 0.15
Bidder - Timestamp: 17, User ID: 8, Action: BID, Loan ID: 430, Rate: 0.10 -- miss
Bidder - Timestamp: 18, User ID: 1, Action: BID, Loan ID: 431, Rate: 0.15
Bidder - Timestamp: 19, User ID: 3, Action: BID, Loan ID: 431, Rate: 0.14
Bidder - Timestamp: 21, User ID: 3, Action: BID, Loan ID: 431, Rate: 0.14

我正在尝试查找所有匹配的User ID，但是这样做很麻烦，我的代码遇到的问题是我最终错过了上面显示的一个匹配的ID。 / p>

这是我当前拥有的代码，该代码获取第一个元素并将其与下一个元素进行比较：

self._bidders = [] # List of bidders

for idx, firstElement in enumerate(self._bidders):
    FirstElement = firstElement
    NextElement = self._bidders[(idx + 1) % len(self._bidders)]
    if FirstElement.user_id == NextElement.user_id:
       #Do something

我将如何确保我得到所有匹配的User ID而不会丢失任何内容且未使用任何导入？，任何建议/帮助将不胜感激。

Answer 1

这是一种将对象列表转换为数据框的方法，您可以从中轻松找到匹配项：

# Create a list of lists
data = ['Bidder - Timestamp: 11, User ID: 8, Action: BID, Loan ID: 430, Rate: 0.20',
        'Bidder - Timestamp: 13, User ID: 13, Action: BID, Loan ID: 430, Rate: 0.15',
        'Bidder - Timestamp: 17, User ID: 8, Action: BID, Loan ID: 430, Rate: 0.10',
        'Bidder - Timestamp: 18, User ID: 1, Action: BID, Loan ID: 431, Rate: 0.15',
        'Bidder - Timestamp: 19, User ID: 3, Action: BID, Loan ID: 431, Rate: 0.14',
        'Bidder - Timestamp: 21, User ID: 3, Action: BID, Loan ID: 431, Rate: 0.14']

df = pd.DataFrame([d.split(',') for d in data])
# df = pd.DataFrame([list(d).split(',') for d in data]) # Use this for your list of objects

df2 = pd.DataFrame()
for i in range(len(df.columns)):
    name = df.iloc[:,i].str.split(':', expand=True)[0][0].strip()
    values = df.iloc[:,i].str.split(':', expand=True)[1].str.strip()
    df2[name] = values

print(df2)

  Bidder - Timestamp User ID Action Loan ID  Rate
0                 11       8    BID     430  0.20
1                 13      13    BID     430  0.15
2                 17       8    BID     430  0.10
3                 18       1    BID     431  0.15
4                 19       3    BID     431  0.14
5                 21       3    BID     431  0.14

# Find matches
df2[df2['User ID'] == '8']

  Bidder - Timestamp User ID Action Loan ID  Rate
0                 11       8    BID     430  0.20
2                 17       8    BID     430  0.10

Answer 2

尝试一下，一个for循环：

vect = TfidfVectorizer()
X = vect.fit_transform(text_list)
word_list = vect.get_feature_names()

df1 = pd.DataFrame(X.toarray())
df1.to_excel("temp1.xlsx")

df2 = pd.DataFrame(X.toarray(), columns = word_list)
df2.to_excel("temp2.xlsx")

或者是单线列表理解：

data = ['Bidder - Timestamp: 11, User ID: 8, Action: BID, Loan ID: 430, Rate: 0.20',
        'Bidder - Timestamp: 13, User ID: 13, Action: BID, Loan ID: 430, Rate: 0.15',
        'Bidder - Timestamp: 17, User ID: 8, Action: BID, Loan ID: 430, Rate: 0.10',
        'Bidder - Timestamp: 18, User ID: 1, Action: BID, Loan ID: 431, Rate: 0.15',
        'Bidder - Timestamp: 19, User ID: 3, Action: BID, Loan ID: 431, Rate: 0.14',
        'Bidder - Timestamp: 21, User ID: 3, Action: BID, Loan ID: 431, Rate: 0.14']
l = [[i.split(':')[0].strip() for i in data[0].split(',')]]
for i in data:
    l.append([x.split(':')[1].strip() for x in i.split(',')])

Answer 3

您可以使用dictionary首先存储ID在列表中出现的次数。然后，您可以根据ID是否重复来过滤原始列表：

bidders = [
    Bidder(11, 8, 'BID', 430, 0.20),
    Bidder(13, 13, 'BID', 430, 0.15),
    Bidder(17, 8, 'BID', 430, 0.10),
    Bidder(18, 1, 'BID', 431, 0.15),
    Bidder(19, 3, 'BID', 431, 0.14),
    Bidder(21, 3, 'BID', 431, 0.14)
]

id_counts = {}
for b in bidders:
    if b.user_id in id_counts:
        id_counts[b.user_id] += 1
    else:
        id_counts[b.user_id] = 1

result = [b for b in bidders if id_counts[b.user_id] > 1]
print(result)

输出

[Timestamp: 11, User Id: 8, Action: BID, Loan ID: 430, Rate: 0.2,
 Timestamp: 17, User Id: 8, Action: BID, Loan ID: 430, Rate: 0.1,
 Timestamp: 19, User Id: 3, Action: BID, Loan ID: 431, Rate: 0.14,
 Timestamp: 21, User Id: 3, Action: BID, Loan ID: 431, Rate: 0.14]

在一个数组中查找匹配的元素

3 个答案: