I have a list similar to this:
host_info = ['stackoverflow.com', '213.213.214.213', 'comments', 'desriprion', 'age']
Sometimes, I need to remove one or more elements from the list and then return the list to the caller. To remove the comments item as an example I'd do this:
host_info.pop(2)
I'd prefer to use a constant for each item rather than some magic number.
Then I could use:
host_info.pop(comments) # comments is integer 2
What's the best way to enumerate these constants so it's simple to maintain them when the list offsets change?
The easiest approach would be to change your list to a dictionary:
keys_list = ['host',...]
host_info = dict(zip(['stackoverflow.com', '213.213.214.213', 'comments', 'desriprion', 'age'], keys_list))
Now you don't have to worry about the key list and if you remove an item from the host_info dictionary it won't affect the key => value associations
or if you only have one host info list:
host_info = {'host':'stackoverflow.com, 'ip':'213.213.214.213',...}
EDIT:
To remove a key from the host_info use pop:
host_info.pop('host')
If you aren't sure the key is there you can use:
host_info.pop('host',None)
or you can use:
if 'host' in host_info:
del host_info['host']
As a side note AdamKG has a nice comment explaining that you wouldn't want to convert this back to a list. You could instead do this:
host = host_info['host']
This removes the dependence on a "key" mapping list and combines the two lists into a single associative container.
Related
I was practicing my grip on dictionaries and then I came across this problem.
I created a dictionary with a few keys and some keys have multiple values which were assigned using lists.
eg:
mypouch = {
'writing_stationery': ['Pens', 'Pencils'],
'gadgets': ['calculator', 'Watch'],
'Documents': ['ID Card', 'Hall Ticket'],
'Misc': ['Eraser', 'Sharpener', 'Sticky Notes']
}
I want to delete a specific item 'Eraser'.
Usually I can just use pop or del function but the element is in a list.
I want the output for misc as 'Misc':['Sharpener', 'Sticky Notes']
What are the possible solutions for this kind of a problem?
You can do:
mypouch['Misc'].remove('Eraser')
Or, use a set rather than a list:
for k in mypouch:
mypouch[k] = set(mypouch[k])
then, it is easy and O(1) to remove an element in a set, using the same code as the list.
Here is a 4 ways to do it the best is with remove but maybe you want to know different approaches
You can use filter
mypouch['Misc'] = list(filter(lambda x: x!='Eraser', mypouch['Misc']))
or short for with if
mypouch['Misc'] = [x for x in mypouch['Misc'] if x != 'Eraser']
or build-in function list remove
mypouch['Misc'].remove('Eraser')
or you can use pop and index combo
mypouch['Misc'].pop(mypouch['Misc'].index('Eraser'))
I am trying to find a way to remove duplicates from a dict list. I don't have to test the entire object contents because the "name" value in a given object is enough to identify duplication (i.e., duplicate name = duplicate object). My current attempt is this;
newResultArray = []
for i in range(0, len(resultArray)):
for j in range(0, len(resultArray)):
if(i != j):
keyI = resultArray[i]['name']
keyJ = resultArray[j]['name']
if(keyI != keyJ):
newResultArray.append(resultArray[i])
, which is wildly incorrect. Grateful for any suggestions. Thank you.
If name is unique, you should just use a dictionary to store your inner dictionaries, with name being the key. Then you won't even have the issue of duplicates, and you can remove from the list in O(1) time.
Since I don't have access to the code that populates resultArray, I'll simply show how you can convert it into a dictionary in linear time. Although the best option would be to use a dictionary instead of resultArray in the first place, if possible.
new_dictionary = {}
for item in resultArray:
new_dictionary[item['name']] = item
If you must have a list in the end, then you can convert back into a dictionary as such:
new_list = [v for k,v in new_dictionary.items()]
Since "name" provides uniqueness... and assuming "name" is a hashable object, you can build an intermediate dictionary keyed by "name". Any like-named dicts will simply overwrite their predecessor in the dict, giving you a list of unique dictionaries.
tmpDict = {result["name"]:result for result in resultArray}
newArray = list(tmpDict.values())
del tmpDict
You could shrink that down to
newArray = list({result["name"]:result for result in resultArray}.values())
which may be a bit obscure.
I have imported a json file and converted it to a python dict. From there I extracted the dict I needed and imported it into a list. Now I would like to manipulate the items further, for example only extract 'player_fullname' : 'Lenny Hampel'. However, it seems that there is another dict inside the list, which len is 1, and I cannot access it.
It is a list with 1 entry and the entry is the dict. Easiest way would be:
player = player[0] #get the dict
print(player['player_fullname'])
Check player_fullname whether in your dict and then compare name value.
[item for item in your_dict if item.get('player_fullname', '') == 'Lenny Hampel']
I have a list like this-
send_recv_pairs = [(['produce_send'], ['consume_recv']), (['Send'], ['Recv']), (['sender2'], ['receiver2'])]
I want something like
[ {['produce_send']:['consume_recv']},{['Send']:['Recv']},{['sender2']:['receiver2']}
How to do this?
You can not use list as the key of dictionary.
This Article explain the concept,
https://wiki.python.org/moin/DictionaryKeys
To be used as a dictionary key, an object must support the hash function (e.g. through hash), equality comparison (e.g. through eq or cmp), and must satisfy the correctness condition above.
And
lists do not provide a valid hash method.
>>> d = {['a']: 1}
TypeError: unhashable type: 'list'
If you want to specifically differentiate the key values you can use tuple as they hash able
{ (i[0][0], ): (i[1][0], ) for i in send_recv_pairs}
{('Send',): ('Recv',),
('produce_send',): ('consume_recv',),
('sender2',): ('receiver2',)}
You can't have lists as keys, only hashable types - strings, numbers, None and such.
If you still want to use a dictionary knowing that, then:
d={}
for tup in send_recv_pairs:
d[tup[0][0]]=tup[1]
If you want the value to be string as well, use tup[1][0] instead of tup[1]
As a one liner:
d={tup[0][0]]:tup[1] for tup in list} #tup[1][0] if you want values as strings
You can check it over here, in the second way of creating distionary.
https://developmentality.wordpress.com/2012/03/30/three-ways-of-creating-dictionaries-in-python/
A Simple way of doing it,
First of all, your tuple is tuple of lists, so better change it to tuple of strings (It makes more sense I guess)
Anyway simple way of working with your current tuple list can be like :
mydict = {}
for i in send_recv_pairs:
print i
mydict[i[0][0]]= i[1][0]
As others pointed out, you cannot use list as key to dictionary. So the term i[0][0] first takes the first element from the tuple - which is a list- and then the first element of list, which is the only element anyway for you.
Do you mean like this?
send_recv_pairs = [(['produce_send'], ['consume_recv']),
(['Send'], ['Recv']),
(['sender2'], ['receiver2'])]
send_recv_dict = {e[0][0]: e[1][0] for e in send_recv_pairs}
Resulting in...
>>> {'produce_send': 'consume_recv', 'Send': 'Recv', 'sender2': 'receiver2'}
As mentioned in other answers, you cannot use a list as a dictionary key as it is not hashable (see links in other answers).
You can therefore just use the values in your lists (assuming they stay as simple as in your example) to create the following two possibilities:
send_recv_pairs = [(['produce_send'], ['consume_recv']), (['Send'], ['Recv']), (['sender2'], ['receiver2'])]
result1 = {}
for t in send_recv_pairs:
result1[t[0][0]] = t[1]
# without any lists
result2 = {}
for t in send_recv_pairs:
result2[t[0][0]] = t[1][0]
Which respectively gives:
>>> result1
{'produce_send': ['consume_recv'], 'Send': ['Recv'], 'sender2': ['receiver2']}
>>> result2
{'produce_send': 'consume_recv', 'Send': 'Recv', 'sender2': 'receiver2'}
Try like this:
res = { x[0]: x[1] for x in pairs } # or x[0][0]: x[1][0] if you wanna store inner values without list-wrapper
It's for Python 3 and when keys are unique. If you need collect list of values per key, instead of single value, than you may use something like itertools.groupby or map+reduce. Wrote about this in comments and I'll provide example.
And yes, list cannot store key-values, only dict's, but maybe it's just typo in question.
You can not use list as the dictionary key, but instead you may type-cast it as tuple to create the dict object.
Below is the sample example using a dictionary comprehension:
>>> send_recv_pairs = [(['produce_send'], ['consume_recv']), (['Send'], ['Recv']), (['sender2'], ['receiver2'])]
>>> {tuple(k): v for k, v in send_recv_pairs}
{('sender2',): ['receiver2'], ('produce_send',): ['consume_recv'], ('Send',): ['Recv']}
For details, take a look at: Why can't I use a list as a dict key in python?
However if your nested tuple pairs were not list, but any other hashable object pairs, you may have type-casted it to dict for getting the desired result. For example:
>>> my_list = [('key1', 'value1'), ('key2', 'value2')]
>>> dict(my_list)
{'key1': 'value1', 'key2': 'value2'}
the current code I have is category1[name]=(number) however if the same name comes up the value in the dictionary is replaced by the new number how would I make it so instead of the value being replaced the original value is kept and the new value is also added, giving the key two values now, thanks.
You would have to make the dictionary point to lists instead of numbers, for example if you had two numbers for category cat1:
categories["cat1"] = [21, 78]
To make sure you add the new numbers to the list rather than replacing them, check it's in there first before adding it:
cat_val = # Some value
if cat_key in categories:
categories[cat_key].append(cat_val)
else:
# Initialise it to a list containing one item
categories[cat_key] = [cat_val]
To access the values, you simply use categories[cat_key] which would return [12] if there was one key with the value 12, and [12, 95] if there were two values for that key.
Note that if you don't want to store duplicate keys you can use a set rather than a list:
cat_val = # Some value
if cat_key in categories:
categories[cat_key].add(cat_val)
else:
# Initialise it to a set containing one item
categories[cat_key] = set(cat_val)
a key only has one value, you would need to make the value a tuple or list etc
If you know you are going to have multiple values for a key then i suggest you make the values capable of handling this when they are created
It's a little hard to understand your question.
I think you want this:
>>> d[key] = [4]
>>> d[key].append(5)
>>> d[key]
[4, 5]
Depending on what you expect, you could check if name - a key in your dictionary - already exists. If so, you might be able to change its current value to a list, containing both the previous and the new value.
I didn't test this, but maybe you want something like this:
mydict = {'key_1' : 'value_1', 'key_2' : 'value_2'}
another_key = 'key_2'
another_value = 'value_3'
if another_key in mydict.keys():
# another_key does already exist in mydict
mydict[another_key] = [mydict[another_key], another_value]
else:
# another_key doesn't exist in mydict
mydict[another_key] = another_value
Be careful when doing this more than one time! If it could happen that you want to store more than two values, you might want to add another check - to see if mydict[another_key] already is a list. If so, use .append() to add the third, fourth, ... value to it.
Otherwise you would get a collection of nested lists.
You can create a dictionary in which you map a key to a list of values, in which you would want to append a new value to the lists of values stored at each key.
d = dict([])
d["name"] = 1
x = d["name"]
d["name"] = [1] + x
I guess this is the easiest way:
category1 = {}
category1['firstKey'] = [7]
category1['firstKey'] += [9]
category1['firstKey']
should give you:
[7, 9]
So, just use lists of numbers instead of numbers.