词典字典，获取值的通用键

Question

I have a very big dictionary of dictionaries and it is like this:

Dict={,
    ...
    '25465466':{'Cmstrk': 'cms_trk_dcs_05:CAEN', 'Crate': 'easyCrate0', 'Board': 'easyBoard06', 'Branch': 'branchController05', 'TrackerSY': 'CMS_TRACKER_SY1527_4', 'Channel': 'channel003\n'},
    '436232302': {'Cmstrk': 'cms_trk_dcs_03:CAEN', 'Crate': 'easyCrate1', 'Board': 'easyBoard01', 'Branch': 'branchController05', 'TrackerSY': 'CMS_TRACKER_SY1527_8', 'Channel': 'channel002\n'},
    '470311412': {'Cmstrk': 'cms_trk_dcs_03:CAEN', 'Crate': 'easyCrate0', 'Board': 'easyBoard00', 'Branch': 'branchController05', 'TrackerSY': 'CMS_TRACKER_SY1527_4', 'Channel': 'channel003\n'},
    ...
} 

And if the user types cms_trk_dcs_05:CAEN or easyCrate1/easyBoard01 or more combination of this values the script has to return those keys (numbers like '654546536') that they have in common.

For example if input is easyCrate0/CMS_TRACKER_SY1527_4 the answer is 470311412,25465466 .

Answer 1

I guess this should help.

Dict = {
'25465466':{'Cmstrk': 'cms_trk_dcs_05:CAEN', 'Crate': 'easyCrate0', 'Board': 'easyBoard06', 'Branch': 'branchController05', 'TrackerSY': 'CMS_TRACKER_SY1527_4', 'Channel': 'channel003\n'},

'436232302': {'Cmstrk': 'cms_trk_dcs_03:CAEN', 'Crate': 'easyCrate1', 'Board': 'easyBoard01', 'Branch': 'branchController05', 'TrackerSY': 'CMS_TRACKER_SY1527_8', 'Channel': 'channel002\n'},

'470311412': {'Cmstrk': 'cms_trk_dcs_03:CAEN', 'Crate': 'easyCrate0', 'Board': 'easyBoard00', 'Branch': 'branchController05', 'TrackerSY': 'CMS_TRACKER_SY1527_4', 'Channel': 'channel003\n'}
}

def sub_search(terms):
    check = lambda k: all(map(lambda term:term in Dict[k].values(), terms))
    return [key for key,value in Dict.items() if check(key)]

def search(term):
    terms = term.split('/')
    return sub_search(terms)

search('cms_trk_dcs_05:CAEN')
#['25465466']

search('easyCrate1/easyBoard01')
#['436232302']

Answer 2

inp = "easyCrate0/CMS_TRACKER_SY1527_4".split("/")
common = []
for k in Dict:
    if all(x in Dict[k].values() for x in inp): # if all of the values match the user input, add it to common list
        common.append(k) 
print (common)
['25465466', '470311412']

You could also use sets:

common = []
for k in Dict:
    if len(set(inp).intersection(Dict[k].values())) == len(inp):
        common.append(k)

You can use a list comp also:

common = [k  for k in Dict if len(set(inp).intersection(Dict[k].values())) == len(inp)]

Answer 3

FINAL EDIT: Here is the code cleaner:

from collections import OrderedDict

find = raw_input("What are you looking for? ")
finder = find.split('/')

similar = []
for key in Dict:
    for x in range(0, len(finder)):
        for pairs in Dict[key].items():
            if pairs[1] == finder[0]: 
                similar.append(key)

if len(similar) == 1 or len(finder) == 1:
    print ",".join(similar) 
for items in similar:
    similar.remove(items)
print ",".join(list(OrderedDict.fromkeys(similar)))