python 3 map/lambda method with 2 inputs

Question

I have a dictionary like the following in python 3:

ss = {'a':'2', 'b','3'}

I want to convert all he values to int using map function, and I wrote something like this:

list(map(lambda key,val: int(val), ss.items())))

but the python complains:

TypeError: () missing 1 required positional argument: 'val'

My question is how can I write a lambda function with two inputs (E.g. key and val)

You've written it with two parameters just fine, but the problem is it is only being called with a single argument, a tuple (key, val). — jonrsharpe
– jonrsharpe, Commented Oct 24, 2014 at 7:31
If you don't want to do anything with the keys, why don't you just map on ss.values()? — BrenBarn
– BrenBarn, Commented Oct 24, 2014 at 7:38
You need to use itertools.starmap instead of plain map as it unpacks the arguments for us(but only if you're using both keys and values, otherwise it's unnecessary as BrenBarn pointed out). In Python 2 it was possible using simple tuple argument unpacking: lambda (key, val): int(val). — Ashwini Chaudhary
– Ashwini Chaudhary, Commented Oct 24, 2014 at 7:45
Have you considered dict comprehension: {k: int(v) for k, v in ss.items()} Or if you want to modify the dict inplace: for k, v in ss.items(): ss[k] = int(v) — jfs
– jfs, Commented Oct 24, 2014 at 14:01
If you need only values then: values = map(int, ss.values()) — jfs
– jfs, Commented Oct 24, 2014 at 14:04

Community · Accepted Answer · 2017-05-23 12:25:13Z

30

ss.items() will give an iterable, which gives tuples on every iteration. In your lambda function, you have defined it to accept two parameters, but the tuple will be treated as a single argument. So there is no value to be passed to the second parameter.

You can fix it like this

print(list(map(lambda args: int(args[1]), ss.items())))
# [3, 2]

If you are ignoring the keys anyway, simply use ss.values() like this
```
print(list(map(int, ss.values())))
# [3, 2]
```

Otherwise, as suggested by Ashwini Chaudhary, using itertools.starmap,

from itertools import starmap
print(list(starmap(lambda key, value: int(value), ss.items())))
# [3, 2]

I would prefer the List comprehension way

print([int(value) for value in ss.values()])
# [3, 2]

In Python 2.x, you could have done that like this

print map(lambda (key, value): int(value), ss.items())

This feature is called Tuple parameter unpacking. But this is removed in Python 3.x. Read more about it in PEP-3113

edited May 23, 2017 at 12:25

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answered Oct 24, 2014 at 7:39

thefourtheye

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3 Comments

Jon Clements Over a year ago

Any other ways you fancy mentioning? :p

jfs Over a year ago

Option zero is to call ss.values(): result = map(int, ss.values())

Mohammadreza Over a year ago

Thank you for the holistic response. also thanks Ashwini for starmaps. I didnt know that difference in python 2 and 3. which was my main problem.

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