Python: Generating a new name based on value in multiple dictionary












0















Relatively new to python, apologies if the question/script is verbose and vague. The script below returns the key and value of the team name searched (Ex: 2, Charlotte Hornets); however, I would also like it to return the league name it's in as well. Such as 'nfl', 'nba', 'mlb'; based on the dictionary it's found in (Ex: 2, Charlotte Hornets, nba). Been working on this for a while, but can't find a dynamic solution. Thanks for the help in advance!



NFL_Teams = {1: "Arizona Cardinals", 2: "Atlanta Falcons", 3: "Baltimore Ravens"}
NBA_Teams = {1:'Washington Wizards', 2:'Charlotte Hornets', 3: 'Atlanta Hawks'}
MLB_Teams = {1: 'Los Angeles Dogers', 2: 'Cincinnati Reds', 3: 'Toronto Blue Jays'}

def Standings(reply):
def dictionary_search(name, dictionary):
for key, value in dictionary.items():
if value == name:
return True # Boolean to show if team name is in merged dictionaries.
if key == name:
pass # Used as a throw-away variable

for single_dictionary in (NFL_Teams, NBA_Teams, MLB_Teams):
if dictionary_search(reply, single_dictionary):
for key, value in single_dictionary.items():
if reply == value:
print(key, value)
break
else:
print('Please Try Again')

Standings('Charlotte Hornets')









share|improve this question


















  • 1





    You basically want to get the name of the variable that holds the data. You can't do this (well, not without some weird trickery), but you can just make a dictionary of dictionaries, where the keys would be the names of the teams, and the values - your original dictionaries (which can be simply lists, BTW).

    – ForceBru
    Dec 30 '18 at 20:22
















0















Relatively new to python, apologies if the question/script is verbose and vague. The script below returns the key and value of the team name searched (Ex: 2, Charlotte Hornets); however, I would also like it to return the league name it's in as well. Such as 'nfl', 'nba', 'mlb'; based on the dictionary it's found in (Ex: 2, Charlotte Hornets, nba). Been working on this for a while, but can't find a dynamic solution. Thanks for the help in advance!



NFL_Teams = {1: "Arizona Cardinals", 2: "Atlanta Falcons", 3: "Baltimore Ravens"}
NBA_Teams = {1:'Washington Wizards', 2:'Charlotte Hornets', 3: 'Atlanta Hawks'}
MLB_Teams = {1: 'Los Angeles Dogers', 2: 'Cincinnati Reds', 3: 'Toronto Blue Jays'}

def Standings(reply):
def dictionary_search(name, dictionary):
for key, value in dictionary.items():
if value == name:
return True # Boolean to show if team name is in merged dictionaries.
if key == name:
pass # Used as a throw-away variable

for single_dictionary in (NFL_Teams, NBA_Teams, MLB_Teams):
if dictionary_search(reply, single_dictionary):
for key, value in single_dictionary.items():
if reply == value:
print(key, value)
break
else:
print('Please Try Again')

Standings('Charlotte Hornets')









share|improve this question


















  • 1





    You basically want to get the name of the variable that holds the data. You can't do this (well, not without some weird trickery), but you can just make a dictionary of dictionaries, where the keys would be the names of the teams, and the values - your original dictionaries (which can be simply lists, BTW).

    – ForceBru
    Dec 30 '18 at 20:22














0












0








0








Relatively new to python, apologies if the question/script is verbose and vague. The script below returns the key and value of the team name searched (Ex: 2, Charlotte Hornets); however, I would also like it to return the league name it's in as well. Such as 'nfl', 'nba', 'mlb'; based on the dictionary it's found in (Ex: 2, Charlotte Hornets, nba). Been working on this for a while, but can't find a dynamic solution. Thanks for the help in advance!



NFL_Teams = {1: "Arizona Cardinals", 2: "Atlanta Falcons", 3: "Baltimore Ravens"}
NBA_Teams = {1:'Washington Wizards', 2:'Charlotte Hornets', 3: 'Atlanta Hawks'}
MLB_Teams = {1: 'Los Angeles Dogers', 2: 'Cincinnati Reds', 3: 'Toronto Blue Jays'}

def Standings(reply):
def dictionary_search(name, dictionary):
for key, value in dictionary.items():
if value == name:
return True # Boolean to show if team name is in merged dictionaries.
if key == name:
pass # Used as a throw-away variable

for single_dictionary in (NFL_Teams, NBA_Teams, MLB_Teams):
if dictionary_search(reply, single_dictionary):
for key, value in single_dictionary.items():
if reply == value:
print(key, value)
break
else:
print('Please Try Again')

Standings('Charlotte Hornets')









share|improve this question














Relatively new to python, apologies if the question/script is verbose and vague. The script below returns the key and value of the team name searched (Ex: 2, Charlotte Hornets); however, I would also like it to return the league name it's in as well. Such as 'nfl', 'nba', 'mlb'; based on the dictionary it's found in (Ex: 2, Charlotte Hornets, nba). Been working on this for a while, but can't find a dynamic solution. Thanks for the help in advance!



NFL_Teams = {1: "Arizona Cardinals", 2: "Atlanta Falcons", 3: "Baltimore Ravens"}
NBA_Teams = {1:'Washington Wizards', 2:'Charlotte Hornets', 3: 'Atlanta Hawks'}
MLB_Teams = {1: 'Los Angeles Dogers', 2: 'Cincinnati Reds', 3: 'Toronto Blue Jays'}

def Standings(reply):
def dictionary_search(name, dictionary):
for key, value in dictionary.items():
if value == name:
return True # Boolean to show if team name is in merged dictionaries.
if key == name:
pass # Used as a throw-away variable

for single_dictionary in (NFL_Teams, NBA_Teams, MLB_Teams):
if dictionary_search(reply, single_dictionary):
for key, value in single_dictionary.items():
if reply == value:
print(key, value)
break
else:
print('Please Try Again')

Standings('Charlotte Hornets')






python python-3.x loops dictionary for-loop






share|improve this question













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asked Dec 30 '18 at 20:18









DG.FinanceDG.Finance

206




206








  • 1





    You basically want to get the name of the variable that holds the data. You can't do this (well, not without some weird trickery), but you can just make a dictionary of dictionaries, where the keys would be the names of the teams, and the values - your original dictionaries (which can be simply lists, BTW).

    – ForceBru
    Dec 30 '18 at 20:22














  • 1





    You basically want to get the name of the variable that holds the data. You can't do this (well, not without some weird trickery), but you can just make a dictionary of dictionaries, where the keys would be the names of the teams, and the values - your original dictionaries (which can be simply lists, BTW).

    – ForceBru
    Dec 30 '18 at 20:22








1




1





You basically want to get the name of the variable that holds the data. You can't do this (well, not without some weird trickery), but you can just make a dictionary of dictionaries, where the keys would be the names of the teams, and the values - your original dictionaries (which can be simply lists, BTW).

– ForceBru
Dec 30 '18 at 20:22





You basically want to get the name of the variable that holds the data. You can't do this (well, not without some weird trickery), but you can just make a dictionary of dictionaries, where the keys would be the names of the teams, and the values - your original dictionaries (which can be simply lists, BTW).

– ForceBru
Dec 30 '18 at 20:22












2 Answers
2






active

oldest

votes


















1














Once you assign the dict to another var called 'single_dictionary', you actually lose the name.. So i tried to map between the league name to the original dict of teams.



Running on:



teams = {'NFL': {1: "Arizona Cardinals", 2: "Atlanta Falcons", 3: "Baltimore Ravens"},
'NBA': {1:'Washington Wizards', 2:'Charlotte Hornets', 3: 'Atlanta Hawks'},
'MLB': {1: 'Los Angeles Dogers', 2: 'Cincinnati Reds', 3: 'Toronto Blue Jays'}}

# NFL_Teams = {1: "Arizona Cardinals", 2: "Atlanta Falcons", 3: "Baltimore Ravens"}
# NBA_Teams = {1:'Washington Wizards', 2:'Charlotte Hornets', 3: 'Atlanta Hawks'}
# MLB_Teams = {1: 'Los Angeles Dogers', 2: 'Cincinnati Reds', 3: 'Toronto Blue Jays'}

def standings(reply):
for league, single_dictionary in teams.items():
if reply in single_dictionary.values():
for key, value in single_dictionary.items():
if reply == value:
print(key, value)
print(league)
break
else:
print('Please Try Again')

standings('Charlotte Hornets')


Will print:



2 Charlotte Hornets
NBA





share|improve this answer
























  • Can't thank you enough! Struggled to think that through for a solid hour.

    – DG.Finance
    Dec 30 '18 at 20:36



















2














Are the numbers used as keys important?
Would it make sense to have read them into the dictionaries the other way around? - then you can use a try except KeyError statement. This might be preferable for very large dictionaries (though this won't be the case here for teams in a league). It also might be a more object orientated way of thinking: the team name is the object identifier, whilst (I presume the numbers are current league positions?) league position is a property (and a changeable one at that). So for example (adapting Aaron_ab's answer)



teams = {'NFL': {"Arizona Cardinals":1, "Atlanta Falcons":2, "Baltimore Ravens":3},
'NBA': {'Washington Wizards':1, 'Charlotte Hornets':2, 'Atlanta Hawks':3},
'MLB': {'Los Angeles Dogers':1, 'Cincinnati Reds':2, 'Toronto Blue Jays':3}}

def standings(team):
for league, teams_dict in teams.items():
try:
teams_dict[team]
print(teams_dict[team], team)
print(league)
break
except KeyError:
continue


Alternatively, drop the numbers and have a dictionary of lists (where the order of the list is the current league order):



import numpy as np

teams = {'NFL': ["Arizona Cardinals", "Atlanta Falcons", "Baltimore Ravens"],
'NBA': ['Washington Wizards', 'Charlotte Hornets', 'Atlanta Hawks'],
'MLB': ['Los Angeles Dogers', 'Cincinnati Reds', 'Toronto Blue Jays']}

def standings(team):
for league, teams_list in teams.items():
if team in teams_list:
print(team, np.where(np.array(teams_list)==team)[0])
print(league)





share|improve this answer
























  • Raising and catching exceptions costs much more computationally wise, try to avoid as much as possible

    – Aaron_ab
    Dec 31 '18 at 7:12











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














Once you assign the dict to another var called 'single_dictionary', you actually lose the name.. So i tried to map between the league name to the original dict of teams.



Running on:



teams = {'NFL': {1: "Arizona Cardinals", 2: "Atlanta Falcons", 3: "Baltimore Ravens"},
'NBA': {1:'Washington Wizards', 2:'Charlotte Hornets', 3: 'Atlanta Hawks'},
'MLB': {1: 'Los Angeles Dogers', 2: 'Cincinnati Reds', 3: 'Toronto Blue Jays'}}

# NFL_Teams = {1: "Arizona Cardinals", 2: "Atlanta Falcons", 3: "Baltimore Ravens"}
# NBA_Teams = {1:'Washington Wizards', 2:'Charlotte Hornets', 3: 'Atlanta Hawks'}
# MLB_Teams = {1: 'Los Angeles Dogers', 2: 'Cincinnati Reds', 3: 'Toronto Blue Jays'}

def standings(reply):
for league, single_dictionary in teams.items():
if reply in single_dictionary.values():
for key, value in single_dictionary.items():
if reply == value:
print(key, value)
print(league)
break
else:
print('Please Try Again')

standings('Charlotte Hornets')


Will print:



2 Charlotte Hornets
NBA





share|improve this answer
























  • Can't thank you enough! Struggled to think that through for a solid hour.

    – DG.Finance
    Dec 30 '18 at 20:36
















1














Once you assign the dict to another var called 'single_dictionary', you actually lose the name.. So i tried to map between the league name to the original dict of teams.



Running on:



teams = {'NFL': {1: "Arizona Cardinals", 2: "Atlanta Falcons", 3: "Baltimore Ravens"},
'NBA': {1:'Washington Wizards', 2:'Charlotte Hornets', 3: 'Atlanta Hawks'},
'MLB': {1: 'Los Angeles Dogers', 2: 'Cincinnati Reds', 3: 'Toronto Blue Jays'}}

# NFL_Teams = {1: "Arizona Cardinals", 2: "Atlanta Falcons", 3: "Baltimore Ravens"}
# NBA_Teams = {1:'Washington Wizards', 2:'Charlotte Hornets', 3: 'Atlanta Hawks'}
# MLB_Teams = {1: 'Los Angeles Dogers', 2: 'Cincinnati Reds', 3: 'Toronto Blue Jays'}

def standings(reply):
for league, single_dictionary in teams.items():
if reply in single_dictionary.values():
for key, value in single_dictionary.items():
if reply == value:
print(key, value)
print(league)
break
else:
print('Please Try Again')

standings('Charlotte Hornets')


Will print:



2 Charlotte Hornets
NBA





share|improve this answer
























  • Can't thank you enough! Struggled to think that through for a solid hour.

    – DG.Finance
    Dec 30 '18 at 20:36














1












1








1







Once you assign the dict to another var called 'single_dictionary', you actually lose the name.. So i tried to map between the league name to the original dict of teams.



Running on:



teams = {'NFL': {1: "Arizona Cardinals", 2: "Atlanta Falcons", 3: "Baltimore Ravens"},
'NBA': {1:'Washington Wizards', 2:'Charlotte Hornets', 3: 'Atlanta Hawks'},
'MLB': {1: 'Los Angeles Dogers', 2: 'Cincinnati Reds', 3: 'Toronto Blue Jays'}}

# NFL_Teams = {1: "Arizona Cardinals", 2: "Atlanta Falcons", 3: "Baltimore Ravens"}
# NBA_Teams = {1:'Washington Wizards', 2:'Charlotte Hornets', 3: 'Atlanta Hawks'}
# MLB_Teams = {1: 'Los Angeles Dogers', 2: 'Cincinnati Reds', 3: 'Toronto Blue Jays'}

def standings(reply):
for league, single_dictionary in teams.items():
if reply in single_dictionary.values():
for key, value in single_dictionary.items():
if reply == value:
print(key, value)
print(league)
break
else:
print('Please Try Again')

standings('Charlotte Hornets')


Will print:



2 Charlotte Hornets
NBA





share|improve this answer













Once you assign the dict to another var called 'single_dictionary', you actually lose the name.. So i tried to map between the league name to the original dict of teams.



Running on:



teams = {'NFL': {1: "Arizona Cardinals", 2: "Atlanta Falcons", 3: "Baltimore Ravens"},
'NBA': {1:'Washington Wizards', 2:'Charlotte Hornets', 3: 'Atlanta Hawks'},
'MLB': {1: 'Los Angeles Dogers', 2: 'Cincinnati Reds', 3: 'Toronto Blue Jays'}}

# NFL_Teams = {1: "Arizona Cardinals", 2: "Atlanta Falcons", 3: "Baltimore Ravens"}
# NBA_Teams = {1:'Washington Wizards', 2:'Charlotte Hornets', 3: 'Atlanta Hawks'}
# MLB_Teams = {1: 'Los Angeles Dogers', 2: 'Cincinnati Reds', 3: 'Toronto Blue Jays'}

def standings(reply):
for league, single_dictionary in teams.items():
if reply in single_dictionary.values():
for key, value in single_dictionary.items():
if reply == value:
print(key, value)
print(league)
break
else:
print('Please Try Again')

standings('Charlotte Hornets')


Will print:



2 Charlotte Hornets
NBA






share|improve this answer












share|improve this answer



share|improve this answer










answered Dec 30 '18 at 20:27









Aaron_abAaron_ab

1,005723




1,005723













  • Can't thank you enough! Struggled to think that through for a solid hour.

    – DG.Finance
    Dec 30 '18 at 20:36



















  • Can't thank you enough! Struggled to think that through for a solid hour.

    – DG.Finance
    Dec 30 '18 at 20:36

















Can't thank you enough! Struggled to think that through for a solid hour.

– DG.Finance
Dec 30 '18 at 20:36





Can't thank you enough! Struggled to think that through for a solid hour.

– DG.Finance
Dec 30 '18 at 20:36













2














Are the numbers used as keys important?
Would it make sense to have read them into the dictionaries the other way around? - then you can use a try except KeyError statement. This might be preferable for very large dictionaries (though this won't be the case here for teams in a league). It also might be a more object orientated way of thinking: the team name is the object identifier, whilst (I presume the numbers are current league positions?) league position is a property (and a changeable one at that). So for example (adapting Aaron_ab's answer)



teams = {'NFL': {"Arizona Cardinals":1, "Atlanta Falcons":2, "Baltimore Ravens":3},
'NBA': {'Washington Wizards':1, 'Charlotte Hornets':2, 'Atlanta Hawks':3},
'MLB': {'Los Angeles Dogers':1, 'Cincinnati Reds':2, 'Toronto Blue Jays':3}}

def standings(team):
for league, teams_dict in teams.items():
try:
teams_dict[team]
print(teams_dict[team], team)
print(league)
break
except KeyError:
continue


Alternatively, drop the numbers and have a dictionary of lists (where the order of the list is the current league order):



import numpy as np

teams = {'NFL': ["Arizona Cardinals", "Atlanta Falcons", "Baltimore Ravens"],
'NBA': ['Washington Wizards', 'Charlotte Hornets', 'Atlanta Hawks'],
'MLB': ['Los Angeles Dogers', 'Cincinnati Reds', 'Toronto Blue Jays']}

def standings(team):
for league, teams_list in teams.items():
if team in teams_list:
print(team, np.where(np.array(teams_list)==team)[0])
print(league)





share|improve this answer
























  • Raising and catching exceptions costs much more computationally wise, try to avoid as much as possible

    – Aaron_ab
    Dec 31 '18 at 7:12
















2














Are the numbers used as keys important?
Would it make sense to have read them into the dictionaries the other way around? - then you can use a try except KeyError statement. This might be preferable for very large dictionaries (though this won't be the case here for teams in a league). It also might be a more object orientated way of thinking: the team name is the object identifier, whilst (I presume the numbers are current league positions?) league position is a property (and a changeable one at that). So for example (adapting Aaron_ab's answer)



teams = {'NFL': {"Arizona Cardinals":1, "Atlanta Falcons":2, "Baltimore Ravens":3},
'NBA': {'Washington Wizards':1, 'Charlotte Hornets':2, 'Atlanta Hawks':3},
'MLB': {'Los Angeles Dogers':1, 'Cincinnati Reds':2, 'Toronto Blue Jays':3}}

def standings(team):
for league, teams_dict in teams.items():
try:
teams_dict[team]
print(teams_dict[team], team)
print(league)
break
except KeyError:
continue


Alternatively, drop the numbers and have a dictionary of lists (where the order of the list is the current league order):



import numpy as np

teams = {'NFL': ["Arizona Cardinals", "Atlanta Falcons", "Baltimore Ravens"],
'NBA': ['Washington Wizards', 'Charlotte Hornets', 'Atlanta Hawks'],
'MLB': ['Los Angeles Dogers', 'Cincinnati Reds', 'Toronto Blue Jays']}

def standings(team):
for league, teams_list in teams.items():
if team in teams_list:
print(team, np.where(np.array(teams_list)==team)[0])
print(league)





share|improve this answer
























  • Raising and catching exceptions costs much more computationally wise, try to avoid as much as possible

    – Aaron_ab
    Dec 31 '18 at 7:12














2












2








2







Are the numbers used as keys important?
Would it make sense to have read them into the dictionaries the other way around? - then you can use a try except KeyError statement. This might be preferable for very large dictionaries (though this won't be the case here for teams in a league). It also might be a more object orientated way of thinking: the team name is the object identifier, whilst (I presume the numbers are current league positions?) league position is a property (and a changeable one at that). So for example (adapting Aaron_ab's answer)



teams = {'NFL': {"Arizona Cardinals":1, "Atlanta Falcons":2, "Baltimore Ravens":3},
'NBA': {'Washington Wizards':1, 'Charlotte Hornets':2, 'Atlanta Hawks':3},
'MLB': {'Los Angeles Dogers':1, 'Cincinnati Reds':2, 'Toronto Blue Jays':3}}

def standings(team):
for league, teams_dict in teams.items():
try:
teams_dict[team]
print(teams_dict[team], team)
print(league)
break
except KeyError:
continue


Alternatively, drop the numbers and have a dictionary of lists (where the order of the list is the current league order):



import numpy as np

teams = {'NFL': ["Arizona Cardinals", "Atlanta Falcons", "Baltimore Ravens"],
'NBA': ['Washington Wizards', 'Charlotte Hornets', 'Atlanta Hawks'],
'MLB': ['Los Angeles Dogers', 'Cincinnati Reds', 'Toronto Blue Jays']}

def standings(team):
for league, teams_list in teams.items():
if team in teams_list:
print(team, np.where(np.array(teams_list)==team)[0])
print(league)





share|improve this answer













Are the numbers used as keys important?
Would it make sense to have read them into the dictionaries the other way around? - then you can use a try except KeyError statement. This might be preferable for very large dictionaries (though this won't be the case here for teams in a league). It also might be a more object orientated way of thinking: the team name is the object identifier, whilst (I presume the numbers are current league positions?) league position is a property (and a changeable one at that). So for example (adapting Aaron_ab's answer)



teams = {'NFL': {"Arizona Cardinals":1, "Atlanta Falcons":2, "Baltimore Ravens":3},
'NBA': {'Washington Wizards':1, 'Charlotte Hornets':2, 'Atlanta Hawks':3},
'MLB': {'Los Angeles Dogers':1, 'Cincinnati Reds':2, 'Toronto Blue Jays':3}}

def standings(team):
for league, teams_dict in teams.items():
try:
teams_dict[team]
print(teams_dict[team], team)
print(league)
break
except KeyError:
continue


Alternatively, drop the numbers and have a dictionary of lists (where the order of the list is the current league order):



import numpy as np

teams = {'NFL': ["Arizona Cardinals", "Atlanta Falcons", "Baltimore Ravens"],
'NBA': ['Washington Wizards', 'Charlotte Hornets', 'Atlanta Hawks'],
'MLB': ['Los Angeles Dogers', 'Cincinnati Reds', 'Toronto Blue Jays']}

def standings(team):
for league, teams_list in teams.items():
if team in teams_list:
print(team, np.where(np.array(teams_list)==team)[0])
print(league)






share|improve this answer












share|improve this answer



share|improve this answer










answered Dec 30 '18 at 22:57









J.WarrenJ.Warren

138119




138119













  • Raising and catching exceptions costs much more computationally wise, try to avoid as much as possible

    – Aaron_ab
    Dec 31 '18 at 7:12



















  • Raising and catching exceptions costs much more computationally wise, try to avoid as much as possible

    – Aaron_ab
    Dec 31 '18 at 7:12

















Raising and catching exceptions costs much more computationally wise, try to avoid as much as possible

– Aaron_ab
Dec 31 '18 at 7:12





Raising and catching exceptions costs much more computationally wise, try to avoid as much as possible

– Aaron_ab
Dec 31 '18 at 7:12


















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