作为索引器提供的不可对齐的布尔系列（布尔系列的索引和索引对象的索引不匹配）。布尔匹配不起作用

Question

filtered = Series([True, False, True], index=df.index)

condition_loc = df.loc[df. LoanAmount.head() < 500]

boolean_i = df.iloc[[True , False ,  True ]]

boolean = df.loc[['True' , 'False' , 'True' ]].values

产生错误

IndexError: Boolean index has wrong length: 3 instead of 614

KeyError: "None of [Index(['True', 'False', 'True'], dtype='object', name='Loan_ID')] are in the [index]"

IndexingError(
pandas.core.indexing.IndexingError: Unalignable boolean Series provided as indexer (index of the boolean Series and of the indexed object do not match).

raise ValueError(
ValueError: Length of values (3) does not match length of index (614)

数据快照

    Loan_ID Gender Married  Dependents     Education Self_Employed  ApplicantIncome  CoapplicantIncome  LoanAmount  Loan_Amount_Term  Credit_History Property_Area Loan_Status
0  LP001002   Male      No           0      Graduate            No             5849                  0         100               360               1         Urban           Y
1  LP001003   Male     Yes           1      Graduate            No             4583               1508         128               360               1         Rural           N
2  LP001005   Male     Yes           0      Graduate           Yes             3000                  0          66               360               1         Urban           Y
3  LP001006   Male     Yes           0  Not Graduate            No             2583               2358         120               360               1         Urban           Y

数据为 [614 行 x 12 列] 目的是生成给定的布尔值列表，选择值为 true 的行 已尝试由上述任何错误生成的每个可用链接。 似乎没有人未能使用上述语法生成值。 请将我指向可以解决此问题的链接。 已尝试尽可能多地解释。 大熊猫的新手。 谢谢你的时间！

编辑：

filtered = Series([True, False, True] )

删除索引解决了第一个问题。

编辑 2：

df.loc[Series([True, False, True])]

给

raise IndexingError(
pandas.core.indexing.IndexingError: Unalignable boolean Series provided as indexer (index of the boolean Series and of the indexed object do not match).

建议的链接仅讨论系列，而不讨论如何将其与 loc 或 iloc 结合使用。

编辑 3：

import pandas as pd 
mydict = [

{"a": 1, "b": 2, "c": 3, "d": 4},

{"a": 100, "b": 200, "c": 300, "d": 400},

{"a": 1000, "b": 2000, "c": 3000, "d": 4000},
]

df = pd.DataFrame(mydict)

print(df)

print(df.iloc[[True, False, True]])

给

a     b     c     d
0     1     2     3     4
1   100   200   300   400
2  1000  2000  3000  4000
      a     b     c     d
0     1     2     3     4
2  1000  2000  3000  4000

适用于上面的代码，其中行等于布尔值但在

print(df.iloc[[True, True]])

编辑 4：

condition_loc = list(filter(lambda x:x.head()>500,df.loc))

给

KeyError: 0
The above exception was the direct cause of the following exception:

    raise KeyError(key) from errKeyError: 0

编辑 5：

boolean = list(compress(loan_df, list1)) 
print(boolean )

打印列名！

编辑 6：

list1 = [True , False ,  True ]
    
boolean = list(compress(df, list1)) 
    for i in boolean :
        print(df.loc[boolean]) 

给

raise KeyError(f"None of [{key}] are in the [{axis_name}]")
KeyError: "None of [Index(['Gender', 'Dependents'], dtype='object', name='Loan_ID')] are in the [index]"

编辑 7： iloc 问题已解决

all_rows_df = list(range(0, len(df))) # gives integer values
    boolean = list(compress(all_rows_df, list1)) # selects values by comparison
    print(boolean)
    for i in boolean :
        print(i)
        print(df.iloc[i]) # Index position of rows in integer or list of integer
        

给

[0, 2]
Gender                   Male
Married                    No
Dependents                  0
Education            Graduate
Self_Employed              No
ApplicantIncome          5849
CoapplicantIncome         0.0
LoanAmount                NaN
Loan_Amount_Term        360.0
Credit_History            1.0
Property_Area           Urban
Loan_Status                 Y
Name: LP001002, dtype: object
Gender                   Male
Married                   Yes
Dependents                  0
Education            Graduate
Self_Employed             Yes
ApplicantIncome          3000
CoapplicantIncome         0.0
LoanAmount               66.0
Loan_Amount_Term        360.0
Credit_History            1.0
Property_Area           Urban
Loan_Status                 Y
Name: LP001005, dtype: object

但是上面的方法在 loc 上报错

[0, 2]
0
KeyError: 0

The above exception was the direct cause of the following exception:

    return self._getitem_axis(maybe_callable, axis=axis)
  
    return self._get_label(key, axis=axis)
  
    return self.obj.xs(label, axis=axis)
  
    loc = index.get_loc(key)
  
    raise KeyError(key) from errKeyError: 0

目前我坚持这个

Answer 1

您需要创建自己的函数以首先转换为字符串，然后拆分并打印回屏幕。

loan_amt = str(loan_df.LoanAmount.head())