How to update value in dictionary?

[ssecorp]

dict.update({"a":1}) SETS the dict item "a" 's value to 1.

i want to increase it by 1. isnt that possible in an easy way? I
should use a tuple for this?

[Iain King]

On Aug 27, 2:40 pm, ssecorp <[e-mail]:

dict.update({"a":1}) SETS the dict item "a" 's value to 1.

i want to increase it by 1. isnt that possible in an easy way? I
should use a tuple for this?

dict["a"] += 1

Iain

[Diez B. Roggisch]

ssecorp wrote:

dict.update({"a":1}) SETS the dict item "a" 's value to 1.

i want to increase it by 1. isnt that possible in an easy way? I
should use a tuple for this?

1) Don't use dict as name for a dictionary, it shadows the type dict

2) setdefault is your friend

d = {}
d['a'] = d.setdefault('a', 1) + 1

Diez

[bearophileHUGS]

ssecorp wrote:

dict.update({"a":1}) SETS the dict item "a" 's value to 1.

That works but it's not the right way to do that, use this:

d["a"] = 1

Bye,
bearophile

[mblume]

Am Wed, 27 Aug 2008 15:45:13 +0200 schrieb Diez B. Roggisch:

>

>dict.update({"a":1}) SETS the dict item "a" 's value to 1.

i want to increase it by 1. isnt that possible in an easy way?
I should use a tuple for this?

1) Don't use dict as name for a dictionary, it shadows the type dict

2) setdefault is your friend

d = {}
d['a'] = d.setdefault('a', 1) + 1

d['a'] = d.get('a', 1) + 1

seems to me a little better, as d['a'] doesn't get set twice, right?

Curious
Martin

[Fredrik Lundh]

mblume wrote:

>2) setdefault is your friend

d = {}
d['a'] = d.setdefault('a', 1) + 1

d['a'] = d.get('a', 1) + 1

seems to me a little better, as d['a'] doesn't get set twice, right?

setdefault is pronounced "get, and set if necessary". it only updates
the dictionary if the key isn't already there:

>>help({}.setdefault)

Help on built-in function setdefault:

setdefault(...)
D.setdefault(k[,d]) -D.get(k,d), also set D[k]=d if k not in D

(since setdefault is a method, the second argument is evaluated whether
it's used or not, but that's true for your approach as well, and isn't
much of a problem for a light-weight object like the integer 1.)

</F>

[MRAB]

On Aug 27, 8:01*pm, Fredrik Lundh <[e-mail]:

mblume wrote:

2) setdefault is your friend

d = {}
d['a'] = d.setdefault('a', 1) + 1

d['a'] = d.get('a', 1) + 1

seems to me a little better, as d['a'] doesn't get set twice, right?

setdefault is pronounced "get, and set if necessary". *it only updates
the dictionary if the key isn't already there:

*>>help({}.setdefault)
Help on built-in function setdefault:

setdefault(...)
* * *D.setdefault(k[,d]) -D.get(k,d), also set D[k]=d if k not in D

(since setdefault is a method, the second argument is evaluated whether
it's used or not, but that's true for your approach as well, and isn't
much of a problem for a light-weight object like the integer 1.)

Both

d['a'] = d.setdefault('a', 1) + 1

and

d['a'] = d.get('a', 1) + 1

will set d['a'] to 2 if 'a' isn't initially in d.