API: how to create a "shallow copy" of a DataFrame?

[jorisvandenbossche]

How can you create a copy of the DataFrame without copying the actual data, but having a new DataFrame that when updated (not in place) does not modify the original ("shallow copy")? And how is this expected to behave?
I suppose that in technical terms, this would be a new BlockManager that references the same arrays?

I ran in the above questions, and actually didn't know a clear answer. The context was: I wanted to replace one column of a DataFrame, but without modifying the original one. And so was wondering if I could do that without making a full copy of the DataFrame (as in theory this is not needed, and I just wanted to update one object column before serializing).

---

So you can do something like this with copy(deep=False). Let's explore this somewhat:

Making a normal (deep) and shallow copy:

In [1]: df = pd.DataFrame({'a': [1, 2, 3], 'b': [.1, .2, .3]}) 

In [2]: df_copy = df.copy() 

In [3]: df_shallow = df.copy(deep=False)

Modifying values in place works as expected: for the copy it does not change the original df, for the shallow copy it does:

In [4]: df_copy.iloc[0,0] = 10  

In [5]: df_shallow.iloc[1,0] = 20  

In [6]: df    
Out[6]: 
    a    b
0   1  0.1
1  20  0.2
2   3  0.3

Overwriting a full column, however, becomes more tricky (due to our BlockManager ...):

# this updates the original df
In [7]: df_shallow['a'] = [10, 20, 30] 

In [8]: df
Out[8]: 
    a    b
0  10  0.1
1  20  0.2
2  30  0.3

# this does not update the original
In [9]: df_shallow['b'] = [100, 200, 300]  

In [10]: df_shallow  
Out[10]: 
    a    b
0  10  100
1  20  200
2  30  300

In [11]: df  
Out[11]: 
    a    b
0  10  0.1
1  20  0.2
2  30  0.3

This is of course somewhat expected if you know the internals: if the new column is of the same dtype, it seems to modify the array of the block in place, while if it needs to create a new block (because the dtype changed on assignment), the reference with the old data is broken and it doesn't modify the original dataframe.

While writing this down, I am realizing that my question is maybe more: should assigning a column (df['a'] = ..) be seen as an in-place modification of your dataframe that has impact through shallow copies?
Because in reality, df['a'] cannot always happen in place (if you are overwriting with a different dtype), this gives rather inconsistent and surprising behaviour depending on the dtypes.

[jorisvandenbossche]

Because in reality, df['a'] cannot always happen in place (if you are overwriting with a different dtype), this gives rather inconsistent and surprising behaviour depending on the dtypes.

So a question could be: should we make this always create a new block, to achieve consistent behaviour? (but of course, this will give a performance degradation in certain cases, as a new consolidation will happen)

Another observation: for extension dtypes, even if it is the same dtype, it always seems to replace the block and not update the values in place:

In [50]: df = pd.DataFrame({'a': pd.array([1, 2, 3], dtype='Int64'), 'b': [.1, .2, .3]})  

In [51]: df_shallow = df.copy(deep=False)  

In [52]: df_shallow['a'] = pd.array([10, 20, 30], dtype='Int64')   

In [53]: df   
Out[53]: 
   a    b
0  1  0.1
1  2  0.2
2  3  0.3

[jsignell]

I don't know about performance implications, but it seems reasonable that you should create a new block every time you assign to an existing column. I think it's best to have a consistent behavior even if that behavior might be slightly less expected.

[jorisvandenbossche]

I agree that consistent behaviour would be nice. But I am not directly sure what possible backward compatibility problems might have.

It's already good that the extension dtypes follow the desired behaviour (maybe we should add an explicit test for that)

[jsignell]

Adding a test sounds like a good idea :)

[jbrockmendel]

With CoW, is obj.copy() now an easy answer for this?

[jorisvandenbossche]

Indeed, obj.copy(deep=False) specifically does now exactly what is described here.