获取NoneType值“ None”的列表

Question

I am new to Python. I'm trying to get a list of NoneType values to create a subplot (plotly library). The following setup is required to create subplots with different specs:

fig = tools.make_subplots(rows=2, cols=3, specs=[ [{'colspan':3}, None, None],
                                                      map(lambda x: {}, ew) ],
                          shared_xaxes=False,   shared_yaxes=False,
                          start_cell='top-left', print_grid=False)

So, based on the values in the list "ew", a list of Nonetype values is needed. The values in the list can vary and so should the list of Nonetypes.

1. Solution: List of Strings, List comprehension:

   lst =', '.join([str(None) for ticker in ew])

   Problem: Strings - convert to Nonetype possible?

2. Solution: Lambda function that inserts None for each value in ew.

   map(lambda x: None, ew)

   Problem: Brackets of the list. Can't get rid of them.

The Solution I am looking for:

print(lst)
None, None
<type 'NoneType'>

so that:

fig = tools.make_subplots(rows=2, cols=3, specs=[ [{'colspan':3}, lst],
                                                      map(lambda x: {}, ew) ],
                          shared_xaxes=False,   shared_yaxes=False,
                          start_cell='top-left', print_grid=False)

Is there a way to get such a list? or a better solution over a embedded function?

EDIT since there is still an error by inserting the following 'lst' to fig:

lst = print(*map(lambda x: None, ew), sep= ', ') #returns None, None 

print('{lst}'.format(**locals())) #returns only None 

-> is this a possible explanation?

Answer 1

You can just create a list and then remove the brackets later.

a = [None, None, None, None, None]  # an example of a list you might want

and you can print it without the brackets like :

print str(a)[1:-1]

Answer 2

If you're using Python 3.x you can try this to print a list without the brackets even if there's a NoneType data included:

print (*lst, sep=', ') #lst = [None, None]

It should output:

None, None

If you're using Python 2.x you can also do this by using from __future__ import print_function

Answer 3

Found an answer for my question: Make one whole list with the dictionary at the first place.

tr = []
for ticker in ew:
    if ew.index(ticker) ==0:
        tr.append({'colspan': len(ew)})
    else:
        tr.append(None)

into fig:

fig = tools.make_subplots(rows=2, cols=3, specs=[ tr, map(lambda x: {},ew)],
                          shared_xaxes=False, shared_yaxes=False,
                          start_cell='top-left', print_grid=False)

more pythonic solutions are welcome.