pandas.DataFrame.append¶

DataFrame.append(other, ignore_index=False, verify_integrity=False, sort=False)[source]¶

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

Deprecated since version 1.4.0: Use concat() instead. For further details see Deprecated DataFrame.append and Series.append

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_indexbool, default False: If True, the resulting axis will be labeled 0, 1, …, n - 1.
verify_integritybool, default False: If True, raise ValueError on creating index with duplicates.
sortbool, default False: Sort columns if the columns of self and other are not aligned.

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

Returns:

DataFrame: A new DataFrame consisting of the rows of caller and the rows of other.

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

pandas.DataFrame.any

pandas.DataFrame.apply