用单行for循环将dict项目附加到列表

Question

I have a list

lst = []

I have dict entries

a= {'a':1,'b':2}

I wish to write a for loop in a comprehension manner filling the list. What I have tried is

lst.append(k,v) for (k,v) in a.items()

I need to then update the dict as

a = {'c':3, 'd':4}

Then again update the list lst .

Which adds the tuples as [('a',1)('b',2)('c',3)('d',4)] What is the right way to iterate through a dict and fill the list?

Answer 1

This is what the syntax for a list comprehension is and should do what you're looking for:

lst = [(k,v) for k,v in a.items()]

In general list comprehension works like this:

someList = [doSomething(x) for x in somethingYouCanIterate]

OUTPUT

>>> lst
[('a', 1), ('b', 2)]

PS Apart from the question asked, you can also get what you're trying to do without list comprehension by simply calling :

lst = a.items()

this will again give you a list of tuples of (key, value) pairs of the dictionary items.

EDIT

After your updated question, since you're updating the dictionary and want the key value pairs in a list, you should do it like:

a= {'a':1,'b':2}
oldA = a.copy()
#after performing some operation
a = {'c':3, 'd':4}
oldA.update(a)
# when all your updates on a is done
lst = oldA.items() #or [(k,v) for k,v in oldA.items()]
# or instead of updating a and maintaining a copy
# you can simply update it like : a.update({'c':3, 'd':4}) instead of a = {'c':3, 'd':4}

Answer 2

One approach is:

a = {"a" : 1, "b" : 2}

lst = [(k, a[k]) for k in a]

a = {"c" : 3, "d" : 4}

lst += [(k, a[k]) for k in a]

Where the contents of lst are [('a', 1), ('b', 2), ('c', 3), ('d', 4)] .

Alternatively, using the dict class' .items() function to accomplish the same:

a = {"a" : 1, "b" : 2}

lst = [b for b in a.items()]

a = {"c" : 3, "d" : 4}

lst += [b for b in a.items()]

Answer 3

There are many valid ways to achieve this. The most easy route is using

a = {"a" : 1, "b" : 2}
lst = list(a.items())

Alternatives include using the zip function, list comprehension etc.