How to pass list of tuple to lamda function

Question

When list of tuples are used in for loop it works perfectly with two separate variables as below

t_dict = {
    "k1": "v1",
    "k2": "v2",
    "k3": "v3",
    "k4": "v4",
    "k5": "v5"
}

for k, v in t_dict.items():
    print "%s=%s" % (k, v)

But when converted into lambda with map function got an error as below

print map(lambda k, v: "%s=%s" % (k, v), t_dict.items())

Traceback (most recent call last):   File "test.py", line 14, in <module>
    print map(lambda k, v: "%s=%s" % (k, v), t_dict.items()) TypeError: <lambda>() takes exactly 2 arguments (1 given)

Is there any other way to call list of tuple in lambda function?

score 3 · Accepted Answer · answered Jan 14 '19 at 09:55

3

Built-in map supports multiple iterables:

res = map(lambda k, v: "%s=%s" % (k, v), t_dict, t_dict.values())
# ['k1=v1', 'k2=v2', 'k3=v3', 'k4=v4', 'k5=v5']

As described in the docs for map:

If additional iterable arguments are passed, function must take that many arguments and is applied to the items from all iterables in parallel.

answered Jan 14 '19 at 09:55

jpp

159,742
34
281
339

there should be a more elegant way, `map('{}={}'.format, t_dict, t_dict.values())` is the best I could think of – Chris_Rands Jan 14 '19 at 10:06
@Chris_Rands, Yup, good point. However, I wanted to advise OP on the more general issue of how to use a multi-argument function with `map`. – jpp Jan 14 '19 at 10:07

score 1 · Answer 2 · answered Jan 14 '19 at 09:59

1

For you case, you can also use the tuple after the % for formatting, so:

map(lambda t: "%s=%s" % t, t_dict.items())

answered Jan 14 '19 at 09:59

mfrackowiak

1,294
8
11

tobias_k · Answer 3 · 2019-01-14T10:19:45.693

1

As already suggested, you can pass multiple iterables to map, but if you want to pass the items and not keys and values individually, you can use zip(*...) to "transpose" the items to two lists and use * again to pass those as two different arguments to map:

>>> list(map(lambda k, v: "%s=%s" % (k, v), *zip(*t_dict.items())))
['k1=v1', 'k2=v2', 'k3=v3', 'k4=v4', 'k5=v5']

Or use itertools.starmap:

>>> from itertools import starmap
>>> list(starmap(lambda k, v: "%s=%s" % (k, v), t_dict.items()))
['k1=v1', 'k2=v2', 'k3=v3', 'k4=v4', 'k5=v5']

edited Jan 14 '19 at 10:19

answered Jan 14 '19 at 10:08

tobias_k

81,265
12
120
179

1

@Chris_Rands Yes, that's nicer that the `lambda`, but it's still a function taking two paramters, thus this does not really change anything about the core issue. – tobias_k Jan 14 '19 at 10:12

iGian · Answer 4 · 2019-01-14T14:44:12.453

0

Other option is this way (str.join(iterable)) to get the list of strings:

map( lambda t: "=".join(t), t_dict.items() )
#=> ['k3=v3', 'k2=v2', 'k1=v1', 'k5=v5', 'k4=v4']

This version can also print out:

import sys
map( lambda t: sys.stdout.write("=".join(t) + "\n"), t_dict.items() )

# k3=v3
# k2=v2
# k1=v1
# k5=v5
# k4=v4

edited Jan 14 '19 at 14:44

answered Jan 14 '19 at 14:37

iGian

11,023
3
21
36

How to pass list of tuple to lamda function

4 Answers4