With regular Python lists, I can do the following:
x = [1, 2, 3]
y = x
x[:] = [4, 5, 6]
print(y) # prints '[4, 5, 6]'
That is, I can replace the content of a list object (x) in-place with completely different values without loosing previously created references to it (y).
I'd like to do the same with a Pandas DataFrame. The following works, but only if the replacement has the same shape and columns as the original:
x = pd.DataFrame({'a': [1,2], 'b': [3,4]})
y = x
x[:] = pd.DataFrame({'a': [5, 6], 'b': [7, 8]})
print(x)
print(y)
If the replacement has different column names then they are silently mapped to the original ones:
x[:] = pd.DataFrame({'c': [9, 10], 'd': [11, 12]})
print(x)
And if the shape of the replacement differs then a ValueError is raised:
x[:] = pd.DataFrame({
'a': [13, 14, 15],
'b': [16, 17, 18],
'e': [18, 19, 20],
})
