pandas.DataFrame.append

DataFrame.append(self, other, ignore_index=False, verify_integrity=False, sort=None)[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
otherDataFrame or Series/dict-like object, or list of these

The data to append.

ignore_indexboolean, default False

If True, do not use the index labels.

verify_integrityboolean, default False

If True, raise ValueError on creating index with duplicates.

sortboolean, default None

Sort columns if the columns of self and other are not aligned. The default sorting is deprecated and will change to not-sorting in a future version of pandas. Explicitly pass sort=True to silence the warning and sort. Explicitly pass sort=False to silence the warning and not sort.

New in version 0.23.0.

Returns
DataFrame

See also

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
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