I tried doing this:
def func(dict):
if dict[a] == dict[b]:
dict[c] = dict[a]
return dict
num = { "a": 1, "b": 2, "c": 2}
print(func(**num))
But it gives in TypeError. Func got an unexpected argument a
3 Answers
Using ** will unpack the dictionary, in your case you should just pass a reference to num to func, i.e.
print(func(num))
(Unpacking ** is the equivalent of func(a = 1, b = 2, c = 3)), e.g.
def func(arg1, arg2):
return arg1 + arg2
args = {"arg1": 3,"arg2": 4}
print(func(**args))
1 Comment
pfabri
The first one should be
print(func(num)), shouldn't it?Two main problems:
- passing the ** argument is incorrect. Just pass the dictionary; Python will handle the referencing.
- you tried to reference locations with uninitialized variable names. Letters a/b/c are literal strings in your dictionary.
Code:
def func(table):
if table['a'] == table['b']:
table['c'] = table['a']
return table
num = { "a": 1, "b": 2, "c": 2}
print(func(num))
Now, let's try a couple of test cases: one with a & b different, one matching:
>>> letter_count = { "a": 1, "b": 2, "c": 2}
>>> print(func(letter_count))
{'b': 2, 'c': 2, 'a': 1}
>>> letter_count = { "a": 1, "b": 1, "c": 2}
>>> print(func(letter_count))
{'b': 1, 'c': 1, 'a': 1}
1 Comment
Dimitris Fasarakis Hilliard
Also, it would be good to point out that using
dict for a name is usually a very silly decision.You can do it like this i don't know why it works but it just work
gender_ = {'Male': 0, 'Female': 1}
def number_wordy(row, **dic):
for key, value in dic.items() :
if row == str(key) :
return value
print(number_wordy('Female', **gender_ ))
1 Comment
whale_steward
hi, eman please add space/tab in python program, as space/tab syntax is important in python.
Iftoif. Second, do not usedictas a parameter or variable name, it is a builtin. Third, why are you sayingfunc(**num)instead offunc(num)?