Question

我习惯于为我所做的任何熊猫操作/更改编写单行代码，而当我回到它之后，我很难阅读/理解（类似于编写一条长的SQL语句。）。有哪些方法可以提高链接操作的可读性？现在我正在尝试类似的东西：

res = (
    # (1) we filter on new__status_group = 'Unknown'
    df[df['new__status_group'] == 'UNKNOWN']

    # (2) we only care about these two files
    [['new__status', 'file_name']]

    # (3) group by the new status
    .groupby('new__status')

    # (4) we want to get the count and value of file_name
    .agg({'file_name': 'first', 'new__status': 'size'})

    # (5) rename the dummy column we used to grab the count
    .rename(columns={'new__status': 'count'})

    # (6) sort the values by count desc
    .sort_values('count', ascending=False)

    # (7) now that we're all good, reset the index so its like a normal data frame with all the fields
    .reset_index()

    # (8) limit to the top ten
    .head(10)

    # (9) and finally we want to pass it as a list of records (dict) for the end usage
    .to_dict('records')

)

这是一个好风格吗？还是这种方式过于笨拙和冗长？还有什么其他方法可以提高熊猫函数的可读性？

Answer 1

两个改进，首先使用loc而不是链片，第二个agg可以传递名称，因此您不需要rename

res = (
    df.loc[df['new__status_group'] == 'UNKNOWN', ['new__status', 'file_name']]
    .groupby('new__status')
    .agg(file_name=('file_name','first'), count=('new__status', 'size'))
    .sort_values('count', ascending=False)
    .reset_index()
    .head(10)
    .to_dict('records')
)

提高熊猫链的可读性

1 个答案: