Problem with dictionary iteration in python

Question

I am new to python so maybe my problem is easy to solve. I have a dictionary organized as follow: my_dict = {'x': (1.1, 3.4), 'y': (4.3, 7.5), 'z': (7.3, 9.5)}. I would like to reduce each value by the same number, so this is what I am trying to do:

for k,v in my_dict.items():
    for a, b in v:
        a = a * 0.3
        b = b * 0.3 
        my_dict[k] = (a, b)
return my_dict

When using this code, I receive this error:

for a, b in v:
TypeError: 'float' object is not iterable

So it seems that the "for a, b in v" loop is trying to iterate through float, which is not possible. Thus, I tried to print my_dict.items() and k,v in the for loop, in order to see what is the for loop iterate through, and this is what I get:

for k,v in my_dict.items():
    print my_dict.items()
    print k,v

[('x', (1.1, 3.4)), ('y', (4.3, 7.5)), ('z', (7.3, 9.5))]
y (4.3, 7.5)

I can see two wierd things: - I was expecting the "print k,v" command to print each key, values couple, but I am getting only the y key with its value - By looking at the my_dict.items(), it seems to me that v is not a float, but a tuple containing two float. So why do I get the float object is not iterable error?

Any help is well appreciated. Again, I am sorry if this is a silly question, but I just started using python and I am stuck with this.

Thank you!

You simply meant a, b = v to unpack the tuple

Thierry Lathuille
– Thierry Lathuille

2020-04-02 17:46:03 +00:00
Commented Apr 2, 2020 at 17:46 — Thierry Lathuille
– Thierry Lathuille, Commented Apr 2, 2020 at 17:46

Cory Kramer · Accepted Answer · 2020-04-02 17:43:45Z

2

You could use a dict comprehension here

>>> {k: tuple(i * 0.3 for i in v) for k, v in my_dict.items()}
{'x': (0.33, 1.02), 'y': (1.2899999999999998, 2.25), 'z': (2.19, 2.85)}

answered Apr 2, 2020 at 17:43

Cory Kramer

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3 Comments

ginopino Over a year ago

This sound great! Thanks! What if I want to make a * 0.3 and b * 0.5?

Prune Over a year ago

In that case, use (v[0]*0.3, v[1]*0.5).

Prune Over a year ago

You could also do it with something more general, such as (i * c for i, c in zip(v, (0.3, 0.5)) for k, v ...)

Prune · Accepted Answer · 2020-04-02 17:48:25Z

1

@CoryKramer already nailed the Pythonic way to do this. To focus on your code ...

Your inner "loop" isn't the right technique. You do not need to iterate through the tuple -- that takes one element at a time, and you want two at once. Rather, you need to "unpack" the tuple.

my_dict = {'x': (1.1, 3.4), 'y': (4.3, 7.5), 'z': (7.3, 9.5)}

for k,v in my_dict.items():
    print("dict:", k, v)
    a, b = v
    print("val:", a, b)
    a = a * 0.3
    b = b * 0.3
    my_dict[k] = (a, b)

answered Apr 2, 2020 at 17:48

Prune

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2 Comments

Anton Pomieshchenko Over a year ago

you can use for k, (a, b) in my_dict.items(): and save one line of code)

Prune Over a year ago

I know -- I'm doing my best to keep close to OP's code.

ericstevens26101 · Accepted Answer · 2020-04-02 17:50:24Z

1

The line of code for a, b in v doesn't make sense because there is only one value v which is your tuple. You are already looping through the items in the dictionary in your outer loop. Try the following:

for k,v in my_dict.items():
    a, b = v
    a = a * 0.3
    b = b * 0.3 
    my_dict[k] = (a, b)
return my_dict

answered Apr 2, 2020 at 17:50

ericstevens26101

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Comments

kederrac · Accepted Answer · 2020-04-02 17:58:18Z

1

you can unpack your dict values:

for k, (a, b) in my_dict.items():
    a *= 0.3
    b *= 0.3 
    my_dict[k] = (a, b)

or you can use:

dict(map(lambda x: (x[0], (x[1][0]*0.3, x[1][1]*0.3)),   my_dict.items()))

answered Apr 2, 2020 at 17:58

kederrac

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Comments

Gabio · Accepted Answer · 2020-04-02 18:08:18Z

1

You don't have to iterate over the tuple but just assign it to variables:

for k,v in my_dict.items():
    a, b = v
    a = a * 0.3
    b = b * 0.3 
    my_dict[k] = (a, b)
return my_dict

and a bit nicer solution:

for k,v in my_dict.items():
    my_dict[k] = tuple([0.3*x for x in v])
return my_dict

edited Apr 2, 2020 at 18:08

answered Apr 2, 2020 at 17:57

Gabio

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Comments

yatu · Accepted Answer · 2020-04-02 18:11:18Z

The problem here, is that by doing:

for a, b in v

You are trying to iterate over a single tuple, which does not really make sense since you're already unpacking it.

Note that all you'd need is to unpack its values as:

k,v = next(iter(my_dict.items()))
a, b = v

Now a and b contains both tuple values respectively. So the only issue in your code is that inner for.

So it is clear that it does not make sense, but why does the error occur?

This is because Python is expecting v to be an iterable of iterables, since you're trying to unpack each iterable into k and v. In this case, when you iterate over v, the first item to be returned is its first float, which cannot be unpacked.

Note that if we instead iterated over a list containing the tuple, your approach would work:

for k,v in my_dict.items():
    for a, b in [v]:
        a = a * 0.3
        b = b * 0.3 
        my_dict[k] = (a, b)

Though, as mentioned it does not make sense, given that you can just directly unpack them:

for k,v in my_dict.items():
    a, b = v
    a = a * 0.3
    b = b * 0.3 
    my_dict[k] = (a, b)

Alain T. · Accepted Answer · 2020-04-02 19:43:57Z

1

I would suggest using a dictionary comprehension:

my_dict = {'x': (1.1, 3.4), 'y': (4.3, 7.5), 'z': (7.3, 9.5)}

f = 0.3
my_dict = {k:(a*f,b*f) for k,(a,b) in my_dict.items()}

print(my_dict)  # {'x': (0.33, 1.02), 'y': (1.2899999999999998, 2.25), 'z': (2.19, 2.85)}

If you want to keep the loop, you will have to remove the nested for-loop because it is not needed to expand the tuples. You can simply write it like this:

for k,(a,b) in my_dict.items():
    my_dict[k] = (a*0.3, b*0.3)

answered Apr 2, 2020 at 19:43

Alain T.

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