python - 为什么 bool 索引返回所有 NaN

Question

我有这个 CSV 文件emp.csv:

index   empno   ename   job mgr hiredate    sal comm    deptno
0,  7839,   KING,   PRESIDENT,  0,  1981-11-17,     5000,   0,  10
1,  7698,   BLAKE,  MANAGER,    7839,   1981-05-01, 2850,   0,  30
2,  7782,   CLARK,  MANAGER,    7839,   1981-05-09, 2450,   0,  10
3,  7566,   JONES,  MANAGER,    7839,   1981-04-01, 2975,   0,  20
4,  7654,   MARTIN, SALESMAN,   7698,   1981-09-10, 1250,   1400,   30
5,  7499,   ALLEN,  SALESMAN,   7698,   1981-02-11, 1600    300,    30
6,  7844,   TURNER, SALESMAN,   7698,   1981-08-21, 1500,   0,  30
7,  7900,   JAMES,  CLERK,      7698,   1981-12-11, 950,    0,  30
8,  7521,   WARD,   SALESMAN,   7698,   1981-02-23, 1250,   500,    30
9,  7902,   FORD,   ANALYST,    7566,   1981-12-11, 3000,   0,  20
10, 7369,   SMITH,  CLERK,      7902,   1980-12-09, 800,    0,  20
11, 7788,   SCOTT,  ANALYST,    7566    1982-12-22, 3000,   0,  20
12, 7876,   ADAMS,  CLERK,      7788,   1983-01-15, 1100,   0,  20
13, 7934,   MILLER, CLERK,      7782,   1982-01-11, 1300,   0,  10

使用下面的代码，我得到了所有的 NaN:

import csv
import sys
import pandas as pd
import dateutil

# Load data from csv file
emp = pd.DataFrame.from_csv("D:\R data\emp.csv")
# Convert date from string to date times`enter code here`
emp['hiredate'] = emp['hiredate'].apply(dateutil.parser.parse, dayfirst=True)
jonessal = emp[['sal']][emp['ename']=='JONES']
empename = emp[['ename','sal']][emp['sal'] > jonessal] 
print(empename)

这是代码的输出:

index           
0       NaN  NaN
1       NaN  NaN
2       NaN  NaN
3       NaN  NaN
4       NaN  NaN
5       NaN  NaN
6       NaN  NaN
7       NaN  NaN
8       NaN  NaN
9       NaN  NaN
10      NaN  NaN
11      NaN  NaN
12      NaN  NaN
13      NaN  NaN

我想要的输出是:

index             
0       KING  5000
9       FORD  3000
11     SCOTT  3000

我认为变量jonesal 的值是2975，但结果是NaN。

如果我用 empename = emp[['ename','sal']][emp['sal'] > 2975 ] 对薪水进行硬编码，它会正常工作，但是当我使用变量代替:empename = emp[['ename','sal']][emp['sal'] > jonessal]

Answer 1

jonessal 是一个 DataFrame。

emp[['ename','sal']][emp['sal'] > jonessal] 

在这里，比较 emp['sal'] > jonessal 不是针对标量的，并且由于广播它返回一个奇怪的 DataFrame。由于索引/形状不匹配，您的最终结果由 NaN 组成。

在这里，您假设只有一名员工名叫琼斯。按照相同的假设，您可以获得标量:

jonessal = emp.loc[emp['ename']=='JONES', 'sal'].values[0]

(.values 返回一个数组，[0] 来自单一员工假设。)

现在，它将返回相同的结果:

emp[['ename','sal']][emp['sal'] > jonessal]
Out[81]: 
    ename   sal
0    KING  5000
9    FORD  3000
11  SCOTT  3000