How to convert dictionary key-value pairs to tuples

Question

Must convert this:

d = {"k1": ("v1"), "k2": ("v2"), "k3": ("v3")}

to this:

[('k1, v1'), ('k2, v2'), ('k2, v3')]

I tried this:

[(k, v) for k, v in d.items()]

But got this:

[('k1', ('v1')), ('k2', ('v2')), ('k3', ('v3'))]

Close, but I cannot have those extra parentheses.

Answer 1

d = {"k1": ("v1"), "k2": ("v2"), "k3": ("v3")}
print [(k, v) for k, v in d.items()]

Already returns

[('k3', 'v3'), ('k2', 'v2'), ('k1', 'v1')]

(which is the same as actually doing list(d.items()) )

---

Now, if your dictionary is actually

d = {"k1": ("v1",), "k2": ("v2",), "k3": ("v3",)}

then that would explain your output. In this case, you need to do

print [(k, v[0]) for k, v in d.items()]

Answer 2

You can try this:

d = {"k1": ("v1"), "k2": ("v2"), "k3": ("v3")}
new_data = list(d.items())

Output:

[('k1', 'v1'), ('k2', 'v2'), ('k3', 'v3')]

Answer 3

Here you go:

>>> d = {"k1": ("v1"), "k2": ("v2"), "k3": ("v3")}
>>> [(a,d[a]) for a in d]
[('k3', 'v3'), ('k2', 'v2'), ('k1', 'v1')]

Answer 4

Just index the string out of the tuple in your list comprehension, ie,

 [(k, v[0]) for k, v in d.items()]

This gives

 [('k3', 'v'), ('k2', 'v'), ('k1', 'v')]