dataframe append 使用方法

叁歲伎倆 2022-12-23 12:25 174阅读 0赞

# pandas.DataFrame.append #

`DataFrame.``append`(*other*, *ignore\_index=False*, *verify\_integrity=False*, *sort=False*)[\[source\]][source]

Append rows of other to the end of caller, returning a new object.

Columns in other that are not in the caller are added as new columns.

Parameters

**other**DataFrame or Series/dict-like object, or list of these

The data to append.

**ignore\_index**bool, default False

If True, the resulting axis will be labeled 0, 1, …, n - 1.

**verify\_integrity**bool, default False

If True, raise ValueError on creating index with duplicates.

**sort**bool, default False

Sort columns if the columns of self and other are not aligned.

New in version 0.23.0.

Changed in version 1.0.0: Changed to not sort by default.

Returns

DataFrame

See also

[`concat`][concat]

General function to concatenate DataFrame or Series objects.

Notes

If a list of dict/series is passed and the keys are all contained in the DataFrame’s index, the order of the columns in the resulting DataFrame will be unchanged.

Iteratively appending rows to a DataFrame can be more computationally intensive than a single concatenate. A better solution is to append those rows to a list and then concatenate the list with the original DataFrame all at once.

Examples

>>> df = pd.DataFrame([[1, 2], [3, 4]], columns=list('AB'))
    >>> df
       A  B
    0  1  2
    1  3  4
    >>> df2 = pd.DataFrame([[5, 6], [7, 8]], columns=list('AB'))
    >>> df.append(df2)
       A  B
    0  1  2
    1  3  4
    0  5  6
    1  7  8

With ignore\_index set to True:

>>> df.append(df2, ignore_index=True)
       A  B
    0  1  2
    1  3  4
    2  5  6
    3  7  8

The following, while not recommended methods for generating DataFrames, show two ways to generate a DataFrame from multiple data sources.

Less efficient:

>>> df = pd.DataFrame(columns=['A'])
    >>> for i in range(5):
    ...     df = df.append({'A': i}, ignore_index=True)
    >>> df
       A
    0  0
    1  1
    2  2
    3  3
    4  4

More efficient:

>>> pd.concat([pd.DataFrame([i], columns=['A']) for i in range(5)],
    ...           ignore_index=True)
       A
    0  0
    1  1
    2  2
    3  3
    4  4

[source]: https://github.com/pandas-dev/pandas/blob/v1.1.4/pandas/core/frame.py#L7614-L7752
[concat]: https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.concat.html#pandas.concat