问题描述

I'm creating a small Pandas dataframe:

df = pd.DataFrame(data={'colA': [["a", "b", "c"]]})

I take a deepcopy of that df. I'm not using the Pandas method but general Python, right?

import copy
df_copy = copy.deepcopy(df)

A df_copy.head() gives the following:

Then I put these values into a dictionary:

mydict = df_copy.to_dict()

That dictionary looks like this:

Finally, I remove one item of the list:

mydict['colA'][0].remove("b")

I'm surprized that the values in df_copy are updated. I'm very confused that the values in the original dataframe are updated too! Both dataframes look like this now:

I understand Pandas doesn't really do deepcopy, but this wasn't a Pandas method. My questions are:

1) how can I build a dictionary from a dataframe that doesn't update the dataframe?

2) how can I take a copy of a dataframe which would be completely independent?

thanks for your help!

Cheers, Nicolas

Disclaimer

Notice that putting mutable objects inside a DataFrame can be an antipattern so make sure that you really need it and you understand what you are doing.

Why doesn't your copy independent

When applied on an object, copy.deepcopy is looked up for a _deepcopy_ method of that object, that is called in turn. It's added to avoid copying too much for objects. In the case of a DataFrame instance in version 0.20.0 and above - _deepcopy_ doesn`t work recursively.

Similarly, if you will use DataFrame.copy(deep=True) deep copy will copy the data, but will not do so recursively. .

How to solve the problem

To take a truly deep copy of a DataFrame containing a list(or other python objects), so that it will be independent - you can use one of the methods below.

df_copy = pd.DataFrame(columns = df.columns, data = copy.deepcopy(df.values))

For a dictionary, you may use same trick:

mydict = pd.DataFrame(columns = df.columns, data = copy.deepcopy(df_copy.values)).to_dict()
mydict['colA'][0].remove("b")

There's also a standard hacky way of deep-copying python objects:

import pickle
df_copy = pickle.loads(pickle.dumps(df))  

Feel free to ask for any clarifications, if needed.

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