Error "'DataFrame' object has no attribute 'append'"

Question

I am trying to append a dictionary to a DataFrame object, but I get the following error:

AttributeError: 'DataFrame' object has no attribute 'append'

As far as I know, DataFrame does have the method "append".

Code snippet:

df = pd.DataFrame(df).append(new_row, ignore_index=True)

I was expecting the dictionary new_row to be added as a new row.

How can I fix it?

Answer 1

As of pandas 2.0, append (previously deprecated) was removed.

You need to use concat instead (for most applications):

df = pd.concat([df, pd.DataFrame([new_row])], ignore_index=True)

As noted by @cottontail, it's also possible to use loc, although this only works if the new index is not already present in the DataFrame (typically, this will be the case if the index is a RangeIndex:

df.loc[len(df)] = new_row # only use with a RangeIndex!

Why was it removed?

We frequently see new users of pandas try to code like they would do it in pure Python. They use iterrows to access items in a loop (see here why you shouldn't), or append in a way that is similar to python list.append.

However, as noted in pandas' issue #35407, pandas's append and list.append are really not the same thing. list.append is in place, while pandas's append creates a new DataFrame:

I think that we should deprecate Series.append and DataFrame.append. They're making an analogy to list.append, but it's a poor analogy since the behavior isn't (and can't be) in place. The data for the index and values needs to be copied to create the result.

These are also apparently popular methods. DataFrame.append is around the 10th most visited page in our API docs.

Unless I'm mistaken, users are always better off building up a list of values and passing them to the constructor, or building up a list of NDFrames followed by a single concat.

As a consequence, while list.append is amortized O(1) at each step of the loop, pandas' append is O(n), making it inefficient when repeated insertion is performed.

What if I need to repeat the process?

Using append or concat repeatedly is not a good idea (this has a quadratic behavior as it creates a new DataFrame for each step).

In such case, the new items should be collected in a list, and at the end of the loop converted to DataFrame and eventually concatenated to the original DataFrame.

lst = []

for new_row in items_generation_logic:
    lst.append(new_row)

# create extension
df_extended = pd.DataFrame(lst, columns=['A', 'B', 'C'])
# or columns=df.columns if identical columns

# concatenate to original
out = pd.concat([df, df_extended])

Answer 2

If it's a single row, loc could also do the job.

df.loc[len(df)] = new_row

With the loc call, the dataframe is enlarged with index label len(df), which makes sense only if the index is RangeIndex; RangeIndex is created by default if an explicit index is not passed to the dataframe constructor.

A working example:

df = pd.DataFrame({'A': range(3), 'B': list('abc')})
df.loc[len(df)] = [4, 'd']
df.loc[len(df)] = {'A': 5, 'B': 'e'}
df.loc[len(df)] = pd.Series({'A': 6, 'B': 'f'})

---

That said, if you are enlarging a dataframe in a loop using DataFrame.append or concat or loc, consider rewriting your code to enlarge a Python list and construct a dataframe once.

As pointed out by @mozway, enlarging a pandas dataframe has O(n^2) complexity because in each iteration, the entire dataframe has to be read and copied. The following perfplot shows the runtime difference relative to concatenation done once.¹ As you can see, both ways to enlarge a dataframe are much, much slower than enlarging a list and constructing a dataframe once (e.g. for a dataframe with 10k rows, concat in a loop is about 800 times slower and loc in a loop is about 1600 times slower).

¹ The code used to produce the perfplot:

import pandas as pd
import perfplot

def concat_loop(lst):
    df = pd.DataFrame(columns=['A', 'B'])
    for dic in lst:
        df = pd.concat([df, pd.DataFrame([dic])], ignore_index=True)
    return df.infer_objects()
    
def concat_once(lst):
    df = pd.DataFrame(columns=['A', 'B'])
    df = pd.concat([df, pd.DataFrame(lst)], ignore_index=True)
    return df.infer_objects()

def loc_loop(lst):
    df = pd.DataFrame(columns=['A', 'B'])
    for dic in lst:
        df.loc[len(df)] = dic
    return df


perfplot.plot(
    setup=lambda n: [{'A': i, 'B': 'a'*(i%5+1)} for i in range(n)],
    kernels=[concat_loop, concat_once, loc_loop],
    labels= ['concat in a loop', 'concat once', 'loc in a loop'],
    n_range=[2**k for k in range(16)],
    xlabel='Length of dataframe',
    title='Enlarging a dataframe in a loop',
    relative_to=1,
    equality_check=pd.DataFrame.equals);

Answer 3

Disclaimer: this answer seems to attract popularity, but the proposed approach should not be used. append was not changed to _append, _append is a private internal method and append was removed from pandas API. The claim "The append method in pandas look similar to list.append in Python. That's why append method in pandas is now modified to _append." is utterly incorrect. The leading _ only means one thing: the method is private and is not intended to be used outside of pandas' internal code.

---

In the new version of Pandas, the append method is changed to _append. You can simply use _append instead of append, i.e., df._append(df2).

df = df1._append(df2,ignore_index=True)

Why is it changed?

The append method in pandas looks similar to list.append in Python. That's why the append method in pandas is now modified to _append.

Answer 4

AttributeError: 'DataFrame' object has no attribute 'append'. Did you mean: '_append'?

For the Pandas with compatibility of TensorFlow 2.12, Users just needs to make a small change for a success.

all_data = train_df.append(test_df)

Change append to _aqppend

all_data = train_df._append(test_df)

And then the program can be run correctly.