python - 没有创建它应该创建的 csv 文件

Question

我想得到一个类似这样的 csv 文件:

index,Basic Mechanics,Basic Thermodynamics,Calculus I,Calculus II,Chemistry I,Chemistry II,Engineering Drawing,Fundamentals of Informatics,Geometry,Linear Algebra
ATTEMPTS,1.416,1.404,1.421,1.347,1.305,1.246,1.261,1.456,1.325,1.385

我有以下代码:

attempts=pd.read_csv("attempts.csv",sep=',', delimiter=None, header='infer')
a2=attempts[(attempts["HSCHOOL"]=="hs1")]
df=a2[["Subject","ATTEMPTS"]]
def subjects_hs(df):
    df=df.pivot_table(columns="Subject",values="ATTEMPTS")
    df=df.reset_index()
    df=df.round(decimals=3)
    df.to_csv("attempts_hs.csv",index=False)

attempts.csv 看起来像这样:

AC_YEAR,HSCHOOL,Subject,ATTEMPTS
2010,hs1,Linear Algebra,2
2010,hs1,Calculus I,1
2010,hs1,Basic Mechanics,2
2010,hs1,Chemistry I,2
2010,hs3,Fundamentals of Informatics,2
2010,hs2,Calculus II,1
2010,hs1,Basic Thermodynamics,1
2010,hs3,Linear Algebra,1
2010,hs3,Calculus I,1
2010,hs2,Basic Mechanics,1
2010,hs2,Chemistry I,1
2010,hs2,Fundamentals of Informatics,1

相反，我得到了以下文件:

Subject,ATTEMPTS
Basic Mechanics,1.545
Basic Thermodynamics,1.667
Calculus I,1.545
Calculus II,1.545
Chemistry I,1.667
Chemistry II,1.4
Engineering Drawing,1.692
Fundamentals of Informatics,2.0
Geometry,1.583
Linear Algebra,1.5

这不是我正在寻找的格式(我想要一行中的所有内容!)。当我正常在 python 中尝试它时，它可以工作，但是在 pythonanywhere 中不是......知道吗？谢谢!

Answer 1

下面的代码似乎对我有用

attempts=pd.read_csv("attempts.csv",sep=',', delimiter=None, header='infer')
a2=attempts[(attempts["HSCHOOL"]=="hs1")]
df=a2[["Subject","ATTEMPTS"]]
def subjects_hs(df):
    df=df.pivot_table(columns="Subject",values=["ATTEMPTS"])
    df=df.reset_index()
    df=df.round(decimals=3)
    df.to_csv("attempts_hs.csv",index=False)

当其“值”参数是列表时，数据透视表提供预期的行为。

第 5 行:将 values="ATTEMPTS" 更改为 values=["ATTEMPTS"]