python可以从列表理解中创建一个字典吗？ [英] Can python create a dictionary from a list comprehension?

问题描述

我正在尝试使用列表将对象列表转换为字典

理解。


类似


entries = {}

[entries [int（d.date.strftime（''％m''）） ）] = d.id]链接中的d]


当我尝试这样做时，我一直都会遇到错误。可能吗？请

字典对象有一个等同于[] .append的方法吗？也许一个

lambda？


感谢您的帮助！

Erik

Hi,

I''m trying to turn o list of objects into a dictionary using a list
comprehension.

Something like

entries = {}
[entries[int(d.date.strftime(''%m''))] = d.id] for d in links]

I keep getting errors when I try to do it. Is it possible? Do
dictionary objects have a method equivalent to [].append? Maybe a
lambda?

Thanks for your help!
Erik

推荐答案

5月27日下午1:55，erikcw< erikwickst ... @ gmail.comwrote：

On May 27, 1:55 pm, erikcw <[e-mail]:

我正在尝试使用列表将对象列表转换为字典

理解。


类似


entries = {}

[entries [int（d.date.strftime（''％m''））] = d.id] for d in links]


当我尝试这样做时，我一直都会遇到错误。可能吗？请

字典对象有一个等同于[] .append的方法吗？也许一个

lambda？


感谢您的帮助！

Erik

Hi,

I''m trying to turn o list of objects into a dictionary using a list
comprehension.

Something like

entries = {}
[entries[int(d.date.strftime(''%m''))] = d.id] for d in links]

I keep getting errors when I try to do it. Is it possible? Do
dictionary objects have a method equivalent to [].append? Maybe a
lambda?

Thanks for your help!
Erik

试试...


[条目.__ setitem __（int（d.date.strftime（''％m''））]，d.id）

链接]


顺便说一下......我也很好奇。我用''dir（dict）''找了一个

的方法，可以做我们想要的和宾果游戏！


~Sean

try...

[entries.__setitem__(int(d.date.strftime(''%m''))], d.id) for d in
links]

btw...I was curious of this too. I used ''dir(dict)'' and looked for a
method that might do what we wanted and bingo!

~Sean

erikcw schrieb：

erikcw schrieb:

我'我试图用列表将对象列表变成字典

理解。


类似


entries = {}

[entries [int（d.date.strftime（''％m''））] = d.id] for d in links]


当我尝试这样做时，我一直都会遇到错误。可能吗？请

字典对象有一个等同于[] .append的方法吗？也许一个

lambda？


感谢您的帮助！

Erik

Hi,

I''m trying to turn o list of objects into a dictionary using a list
comprehension.

Something like

entries = {}
[entries[int(d.date.strftime(''%m''))] = d.id] for d in links]

I keep getting errors when I try to do it. Is it possible? Do
dictionary objects have a method equivalent to [].append? Maybe a
lambda?

Thanks for your help!
Erik

通常是dict（whatEver）会做的;-)


示例：


a = [1,2 ，3,4,5,6,7,8,9,10]

aDict = dict（x中的[（x，x + 1），如果x％2 = = 0]）


打印aDict

normally a dict(whatEver) will do ;-)

Example:

a = [1,2,3,4,5,6,7,8,9,10]

aDict = dict([(x,x+1) for x in a if x%2==0])

print aDict

27 mai，22：55，erikcw< erikwickst。 .. @ gmail.comwrote：

On 27 mai, 22:55, erikcw <[e-mail]:

我正试图把对象列表变成一个字典使用清单

理解。


类似


entries = {}

[entries [int（d.date.strftime（''％m''））] = d.id] for d in links]


我一直遇到错误试着去做。可能吗？请

字典对象有一个等同于[] .append的方法吗？也许一个

lambda？


感谢您的帮助！

Erik

Hi,

I''m trying to turn o list of objects into a dictionary using a list
comprehension.

Something like

entries = {}
[entries[int(d.date.strftime(''%m''))] = d.id] for d in links]

I keep getting errors when I try to do it. Is it possible? Do
dictionary objects have a method equivalent to [].append? Maybe a
lambda?

Thanks for your help!
Erik

entries = dict（[（int（d.date.strftime（''％m''）），d.id）for d in links]）


随着Python2.4及更高版本你可以使用生成器表达式

entries = dict（（int（d.date.strftime（''％m''）），d。 id）for d in links）

问候，

Pierre

entries = dict([ (int(d.date.strftime(''%m'')),d.id) for d in links] )

With Python2.4 and above you can use a "generator expression"

entries = dict( (int(d.date.strftime(''%m'')),d.id) for d in links )
Regards,
Pierre

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