[['Demo-Site', '10.227.209.139'], ['Demo-Site', '10.227.215.68'],
['Demo-Site', '172.18.74.146'], ['Site', '10.152.114.65'],
['Site', '10.227.211.244'], ['Demo-Site', '10.227.147.98'],
['test', '172.18.74.146']]
How can I concatenate all of the IP's to form a big string based on if the first index is the same? Do I make a default dictionary?
Should be:
["Site", "10.227.211.244, 10.152.114.65"]
Here you go, using a defaultdict based solution:
In [1]: list_of_ips = [['Demo-Site', '10.227.209.139'], ['Demo-Site', '10.227.215.68'],
...: ['Demo-Site', '172.18.74.146'], ['test', '10.152.114.65'],
...: ['Site', '10.227.211.244'], ['Demo-Site', '10.227.147.98'],
...: ['test', '172.18.74.146']]
In [2]: from collections import defaultdict
In [3]: resp_dict = defaultdict(list)
In [4]: for item in list_of_ips:
...: resp_dict[item[0]].append(item[1])
...:
In [5]: result = [[key, ", ".join(resp_dict[key])] for key in resp_dict]
In [6]: result
Out[6]:
[['test', '10.152.114.65, 172.18.74.146'],
['Site', '10.227.211.244'],
['Demo-Site', '10.227.209.139, 10.227.215.68, 172.18.74.146, 10.227.147.98']]
You could use normal dictionary setdefault method :
dic={}
for ele in lis:
dic.setdefault(ele[0],[]).append(ele[1])
[[a,','.join(b)] for a,b in dic.items()]
It seems like you want to group your record based on the first element in each sub-list. That is what the groupby does. but there is an important preliminary step which is sorting your list based on the first first element in each sub-list. You can do this using the sorted function and use itemgetter as key function.
from operator import itemgetter
from itertools import groupby
result = []
my_list = [['Demo-Site', '10.227.209.139'],
['Demo-Site', '10.227.215.68'],
['Demo-Site', '172.18.74.146'],
['Site', '10.152.114.65'],
['Site', '10.227.211.244'],
['Demo-Site', '10.227.147.98'],
['test', '172.18.74.146']]
Demo groupby
for g, data in groupby(sorted(my_list, key=itemgetter(0)), itemgetter(0)):
print(g)
for elt in data:
print(' ', elt)
yields:
Demo-Site
['Demo-Site', '10.227.209.139']
['Demo-Site', '10.227.215.68']
['Demo-Site', '172.18.74.146']
['Demo-Site', '10.227.147.98']
Site
['Site', '10.152.114.65']
['Site', '10.227.211.244']
test
['test', '172.18.74.146']
As you can see your data is grouped by first element in your sub-list. So all you need now is concatenate (.join) the last elements of the "members" of the same group and then append a list [<given group>, <members string>] to the result list.
>>> for g, data in groupby(sorted(my_list, key=itemgetter(0)), itemgetter(0)):
... result.append([g, ', '.join(elt[1] for elt in data)])
...
>>> result
[['Demo-Site', '10.227.209.139, 10.227.215.68, 172.18.74.146, 10.227.147.98'], ['Site', '10.152.114.65, 10.227.211.244'], ['test', '172.18.74.146']]
You can use a dictionary (and dict.setdefault method )to preserve your items based on first index,then join the values :
>>> li=[['Demo-Site', '10.227.209.139'], ['Demo-Site', '10.227.215.68'],
... ['Demo-Site', '172.18.74.146'], ['Site', '10.152.114.65'],
... ['Site', '10.227.211.244'], ['Demo-Site', '10.227.147.98'],
... ['test', '172.18.74.146']]
>>>
>>> d={}
>>>
>>> for i,j in li:
... d.setdefault(i,[]).append(j)
...
>>> [[i,','.join(j)] for i,j in d.items()]
[['test', '172.18.74.146'], ['Site', '10.152.114.65,10.227.211.244'], ['Demo-Site', '10.227.209.139,10.227.215.68,172.18.74.146,10.227.147.98']]
>>>
or, as you mentioned, with defaultdict:
from collections import defaultdict
lst = [['Demo-Site', '10.227.209.139'], ['Demo-Site', '10.227.215.68'],
['Demo-Site', '172.18.74.146'], ['Site', '10.152.114.65'],
['Site', '10.227.211.244'], ['Demo-Site', '10.227.147.98'],
['test', '172.18.74.146']]
dct = defaultdict(list)
for name, ip in lst:
dct[name].append(ip)
res = [ [name, ', '.join(ips)] for name, ips in dct.items() ]
print(res)
