print index name in python

Question

I want to know how to output the name, not the values, of each index (and sub-index) in a list and in sub-lists.

For example, let's say I have a main list, which has 3 other lists each containing 5 sub-lists. Here is a small example to clarify:

List1 = [SubList1, SubList2, SubList3, SubList4, SubList5]
List2 = [Apple, Banana, Orange, Pear, Grape]
List3 = [Red, Yellow, Orange, Green, Purple]


Lists = [List1, List2, List3]

MainList = [Lists]

How can I print out 'List1'?

When I type in

MainList[0][0]

it outputs

[SubList1, SubList2, ..., Sublist5]

I want the actual name of the index, not its values.

I thought about using dictionary keys, but the report generator that I am using for the XML conversion outputs all the data into lists. Does this mean I have no choice but to modify my report generator script to output the data into dictionaries?

You can't, the name doesn't work like that. It's a label, not an attribute. Recommended reading: nedbatchelder.com/text/names.html — jonrsharpe
– jonrsharpe, Commented Oct 9, 2015 at 14:36
list index doesn't have name, but key of the dictionaries does have name, use it. — Ahasanul Haque
– Ahasanul Haque, Commented Oct 9, 2015 at 14:37

xiº · Accepted Answer · 2015-10-09 14:42:20Z

2

Try to use dict() for that purpose:

List1 = ['SubList1', 'SubList2', 'SubList3', 'SubList4', 'SubList5']
List2 = ['Apple', 'Banana', 'Orange', 'Pear', 'Grape']
List3 = ['Red', 'Yellow', 'Orange', 'Green', 'Purple']


xd = {'List1': List1, 'List2': List2, 'List3': List3}

print xd['List1']

for k, v in xd.iteritems():
    if v == List1:
        print k

>>> ['SubList1', 'SubList2', 'SubList3', 'SubList4', 'SubList5']
>>> List1

answered Oct 9, 2015 at 14:42

xiº

4,7173 gold badges31 silver badges43 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

Mike Issa Over a year ago

Thank you, I tried your suggestion and I am receiving the following error: AttributeError: 'dict' object has no attribute 'iteritems' I am using Python 3.4 by the way

Mike Issa Over a year ago

Found the answer to my problem: stackoverflow.com/questions/30418481/…

xiº Over a year ago

@MikeIssa yep. You didn't need iteritems() in Python 3. It is generator by default. Just use items().

GingerPlusPlus Over a year ago

Sadly, that's O(n) solution.

GingerPlusPlus · Accepted Answer · 2015-10-09 16:20:26Z

2

You seem to misunderstand. List1 in your example is a variable name. It is not stored in the Lists array, only its value is stored. So you are not going to get List1 out of Lists. Array indexes are integers and the index of List1 in Lists is 0. As the other commenter suggested you might want to convert your Lists structure into a dict.

Lists = {}
Lists['List1'] = [SubList1, SubList2, SubList3, SubList4, SubList5]
Lists['List2'] = [Apple, Banana, Orange, Pear, Grape]
Lists['List3'] = [Red, Yellow, Orange, Green, Purple]

print(Lists.keys())

edited Oct 9, 2015 at 16:20

GingerPlusPlus

5,6743 gold badges31 silver badges54 bronze badges

answered Oct 9, 2015 at 15:16

Mad Wombat

15.3k15 gold badges78 silver badges114 bronze badges

Comments

GingerPlusPlus · Accepted Answer · 2015-10-09 15:16:31Z

1

Quick solution:

class NamedList(list):
    __slots__ = ['__name__'] # optimisation, can be ommited
    def __init__(self, name, *args):
        super().__init__(*args)
        self.__name__ = name
    def __repr__(self):
        return "%s(%r, %s)" % (type(self).__name__, self.__name__, super().__repr__())

test = NamedList('test', [1, 2, 3])
print(test)
print(test.__name__)

Output:

NamedList('test', [1, 2, 3])
test

answered Oct 9, 2015 at 15:16

GingerPlusPlus

5,6743 gold badges31 silver badges54 bronze badges

Comments

G.S · Accepted Answer · 2015-10-09 15:28:34Z

1

list object has no name attributes, you should use pandas.Series as follow:

import pandas as pds
List1 = pds.Series([SubList1, SubList2, SubList3, SubList4, SubList5],name='List1')
List2 = pds.Series([Apple, Banana, Orange, Pear, Grape],name='List2')

you find the name like this:

print List1.name,List2.name

A list of list become a pandas.DataFrame:

Lists = pds.DataFrame([List1, List2, List3])

To answer your question, you should proceed as follow:

-each sublist become a series

Sublist = pds.Series([...],name='Sublist')

-each list become a dataframe:

List = pds.DataFrame([Sublist1,Sublist2,..])

-"the" Lists become a dict:

Lists = {'List1' : List1,
         'List2' : List2,
         ...
         }

-you get the name:

for name_list,list in Lists.items(): 
     name_list,list.columns

edited Oct 9, 2015 at 15:28

answered Oct 9, 2015 at 15:09

G.S

4023 silver badges13 bronze badges

1 Comment

GingerPlusPlus Over a year ago

Please add some example code to show how it solves the problem.

GingerPlusPlus · Accepted Answer · 2015-10-09 15:28:48Z

1

If your lists always are stored as variables, you can retrieve their names by using some sort of identity dict (sadly there is none in standard library):

names = IdentityWeakKeyDict((obj, name for name, obj in locals().items()))

names should now map objects to their names.

answered Oct 9, 2015 at 15:28

GingerPlusPlus

5,6743 gold badges31 silver badges54 bronze badges

Collectives™ on Stack Overflow

print index name in python

5 Answers 5

4 Comments

Comments

Comments

1 Comment

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

4 Comments

Comments

Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related