2

I am trying to generate (something like) the following dictionary:

funcs1 = {
    '0':lambda x:x==0,
    '1':lambda x:x==1,
    '2':lambda x:x==2,
    '3':lambda x:x==3,
    '4':lambda x:x==4,
    '5':lambda x:x==5,
}

I tried to create the dictionary with a list comprehension like this:

funcs2 = {str(i):lambda x:x==i for i in range(0,6)}

Or simply using a for loop:

funcs3 = {}
for i in range(0,6):
    funcs3.update({str(i): lambda x:x==i})

However, funcs2 and funcs3 are not the same as funcs1, for example, when calling the element '0' of each one of them and applying it to 0, the results are different:

funcs1['0'](0)
Out[2]: True

funcs2['0'](0)
Out[3]: False

funcs3['0'](0)
Out[4]: False

Can somebody please help me out and point out where I am making a mistake?

Juan Manuel Ortiz
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2 Answers2

5

This is a common misunderstanding caused by Python late binding, this should fix your code:

funcs1 = {
    '0': lambda x: x == 0,
    '1': lambda x: x == 1,
    '2': lambda x: x == 2,
    '3': lambda x: x == 3,
    '4': lambda x: x == 4,
    '5': lambda x: x == 5,
}

funcs2 = {str(i): lambda x, i=i: x == i for i in range(0, 6)}

funcs3 = {}
for i in range(0, 6):
    funcs3.update({str(i): lambda x, i=i: x == i})

print(funcs1['0'](0))
print(funcs2['0'](0))
print(funcs3['0'](0))

Output

True
True
True
Dani Mesejo
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0

This works:

>>> def make_fn(i):
...     def fn(x):
...         return x==i
...     return fn
...
>>> funcs = {str(i): make_fn(i) for i in range(6)}

Testing, I get:

>>> funcs['0'](0)
True
>>> funcs['1'](1)
True
>>> funcs['0'](1)
False
Job Evers
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