ERR: setting a column with a scalar and no index should raise

[jreback]

xref #10017 (comment)

This should raise like [5] below.

In [2]: df = DataFrame()

In [3]: df['foo'] = 1

In [4]: df
Out[4]: 
Empty DataFrame
Columns: [foo]
Index: []

In [5]: DataFrame({'foo': 1})
ValueError: If using all scalar values, you must pass an index
I
n [17]: df.loc[1] = 1
\ValueError: cannot set a frame with no defined columns

Even though this is technically valid code (in [4]), IOW the scalar is reindexed to a 0-len index, this immediately loses information This should raise an error.

[jreback]

cc @AwasthiMaddy

[toobaz]

I'm not convinced this is a bug... as @jreback says, [3] is valid code, and the boundary with "losing information" is not so obvious. What about

In [2]: s = pd.Series(range(4))

In [3]: s.loc[5:] = 1

? I'm pretty sure we want to allow it. On the other hand, in [5] you are initializing, so it makes sense to require more explicit "instructions" (as in pd.DataFrame({'foo' : 1}, index=[]), which works fine).

I see instead the incoherence with [17], but I think it can be justified (because a DataFrame has a better defined meaning than one with no columns; and because setting an entire column to a scalar is a more usual operation than setting an entire row to a scalar) - and anyway, I would ideally prefer to fix [17] (which should expand the index) than to forbid [3].