Here is my code:
for response in responses["result"]:
ids = {}
key = response['_id'].encode('ascii')
print key
for value in response['docs']:
ids[key].append(value)
Traceback:
File "people.py", line 47, in <module>
ids[key].append(value)
KeyError: 'deanna'
I am trying to add multiple values to a key. Throws an error like above
Check out setdefault:
ids.setdefault(key, []).append(value)
It looks to see if key is in ids, and if not, sets that to be an empty list. Then it returns that list for you to inline call append on.
Docs:
http://docs.python.org/2/library/stdtypes.html#dict.setdefault
If I'm reading this correctly your intention is to map the _id of a response to its docs. In that case you can bring down everything you have above to a dict comprehension:
ids = {response['_id'].encode('ascii'): response['docs']
for response in responses['result']}
This also assumes you meant to have id = {} outside of the outermost loop, but I can't see any other reasonable interpretation.
If the above is not correct,
You can use collections.defaultdict
import collections # at top level
#then in your loop:
ids = collections.defaultdict(list) #instead of ids = {}
A dictionary whose default value will be created by calling the init argument, in this case calling list() will produce an empty list which can then be appended to.
To traverse the dictionary you can iterate over it's items()
for key, val in ids.items():
print(key, val)
The reason you're getting a KeyError is this: In the first iteration of your for loop, you look up the key in an empty dictionary. There is no such key, hence the KeyError.
The code you gave will work, if you first insert an empty list into the dictionary under to appropriate key. Then append the values to the list. Like so:
for response in responses["result"]:
ids = {}
key = response['_id'].encode('ascii')
print key
if key not in ids: ## <-- if we haven't seen key yet
ids[key] = [] ## <-- insert an empty list into the dictionary
for value in response['docs']:
ids[key].append(value)
The previous answers are correct. Both defaultdict and dictionary.setdefault are automatic ways of inserting the empty list.
Related
If I have a list of dictionaries in a python script, that I intend to later on dump in a JSON file as an array of objects, how can I index the keys of a specific dictionary within the list?
Example :
dict_list = [{"first_dict": "some_value"}, {"second_dict":"some_value"}, {"third_dict": "[element1,element2,element3]"}]
My intuitive solution was dict_list[-1][0] (to access the first key of the last dictionary in the list for example). This however gave me the following error:
IndexError: list index out of range
the key inputted into the dictionary will pick the some value in the format dict = {0:some_value}
to find a specific value:
list_dictionary = [{"dict1":'value1'},{"dict2","value2"}]
value1 = list_dictionary[0]["dict1"]
the 'key' is what you have to use to find a value from a dictionary
Example:
dictionary = {0:value}
dictionary[0]
in this case it will work
but to pick the elements we will do
values = []
for dictionary in dict_list:
for element in dictionary:
values.append(dictionary[element])
Output:
['some_value', 'some_value', ['element1', 'element2', 'element3']]
dict_list = [{"first_dict": "some_value"}, {"second_dict":"some_value"}, {"third_dict": ['element1','element2','element3']}]
If your dict look like this you can do as well
dict_list[-1]["third_dict"]
You can't access 'the first key' with a int since you have a dict
You can get the first key with .keys() and then
dict_list[-1].keys()[0]
By using dict_list[-1][0], you are trying to access a list with a list, which you do not have. You have a list with a dict key within a list.
Taking your example dict_list[-1][0]:
When you mention dict_list you are already "in the list".
The first index [-1] is referring to the last item of the list.
The second index would only be "usable" if the item mentioned in the previous index were a list. Hence the error.
Using:
dict_list=[{"first_dict": "some_value"}, {"second_dict":"some_value"},{"third_dict": [0,1,2]}]
to access the value of third_dict you need:
for value in list(dict_list[-1].values())[0]:
print(value)
Output:
0
1
2
If you know the order of dictionary keys and you are using one of the latest python versions (key stays in same order), so:
dict_list = [
{"first_dict": "some_value"}
, {"second_dict":"some_value"}
, {"third_dict": ["element1", "element2", "element3"]}
]
first_key = next(iter(dict_list[-1].keys()))
### OR: value
first_value = next(iter(dict_list[-1].values()))
### OR: both key and value
first_key, first_value = next(iter(dict_list[-1].items()))
print(first_key)
print(first_key, first_value)
print(first_value)
If you have the following list of dictionaries:
dict_list = [{"key1":"val1", "key2":"val2"}, {"key10":"val10"}]
Then to access the last dictionary you'd indeed use dict_list[-1] but this returns a dictionary with is indexed using its keys and not numbers: dict_list[0]["key1"]
To only use numbers, you'd need to get a list of the keys first: list(dict_list[-1]). The first element of this list list(dict_list[-1])[0] would then be the first key "key10"
You can then use indices to access the first key of the last dictionary:
dict_index = -1
key_index = 0
d = dict_list[dict_index]
keys = list(d)
val = d[keys[key_index]]
However you'd be using the dictionary as a list, so maybe a list of lists would be better suited than a list of dictionaries.
i'm trying to append to a dictionary. there are two loops. the name of the keys depends on the value of the inner loop and the key is the value of a variable which is updated within the loop. my script is
def append_value(dict_obj, key, value):
# Check if key exist in dict or not
if key in dict_obj:
# Key exist in dict.
# Check if type of value of key is list or not
if not isinstance(dict_obj[key], list):
# If type is not list then make it list
dict_obj[key] = [dict_obj[key]]
# Append the value in list
dict_obj[key].append(value)
else:
# As key is not in dict,
# so, add key-value pair
dict_obj[key] = value
for x in range(tot):
dict=['output'=x]
for a in range(33,91):
index_val=(a*sum_t)/x
# now i'm trying to create key names that would be year_33 year_34 and so on
head=''
head='year_{}'.format(a)
append_value(dict, head=avg_PMI)
i get the error name 'append_value' is not defined. would appreciate any help. i would like to loop over values of tot and the (33,91) range. each combination of the two gives a unique values and i want to create a dictionary which will become a csv where x values are rows, a is the column.
thanks!
edited: to show append_value function
The cause of the error is that when you call a function with head=avg_PMI, python assumes that head is an argument. For append_value, the only arguments are dict_obj, key and value. I assume that you want to add a value to the dictionary such that head=avg_PMI. To do this, you have to call the function append_value in the following way:
append_value(dict, head, avg_PMI).
I have a default dictionary and I run it through a couple of loops to look for certain strings in the dictionary. The loops don't really append anything to the dictionary yet as it turns out, during the loop, new items keep getting appended to the dictionary and the final dictionary ends up bigger than the original one before the loop.
I've been trying to pinpoint the error forever but now it's late and I have no idea what's causing this!
from collections import defaultdict
dummydict = defaultdict(list)
dummydict['Alex'].append('Naomi and I love hotcakes')
dummydict['Benjamin'].append('Hayley and I hate hotcakes')
part = ['Alex', 'Benjamin', 'Hayley', 'Naomi']
emp = []
for var in dummydict:
if 'I' in dummydict[var]:
emp.append(var)
for car in part:
for key in range(len(dummydict)):
print('new len', len(dummydict))
print(key, dummydict)
if car in dummydict[key]:
emp.append(car)
print(emp)
print('why are there new values in the dictionary?!', len(dummydict), dummydict)
I expect the dictionary to remain unchanged.
if car in dummydict[key]:
key being an integer, and your dict being initially filled with only string as keys, this will create a new value in dummydict for each key.
Accessing missing keys as in dummydict[key] will add those keys to the defaultdict. Note that key is an int, not the value at that position, as for key in range(len(dummydict)) iterates indexes, not the dict or its keys.
See the docs:
When each key is encountered for the first time, it is not already in the mapping; so an entry is automatically created using the default_factory function which returns an empty list.
For example, this code will show a dummydict with a value in it, because simply accessing dummydict[key] will add the key to the dict if that key is not already there.
from collections import defaultdict
dummydict = defaultdict(list)
dummydict[1]
print (dummydict)
outputs:
defaultdict(<class 'list'>, {1: []})
Your issue is that in your loop, you do things like dummydict[key] and dummydict[var], which adds those keys.
I try to append function objects to list, which is element of dictionary:
jobs = {}
job = sched.add_date_job(callback, run_at, [params])
jobs[hereCanBeRandomNumber].append(job)
But, it seems I have a problem in last line. Compiler says:KeyError: 118096950.
What's the problem?
The way you're adding to your dictionary is incorrect.
jobs[hereCanBeRandomNumber].append(job)
Translates to "Append the job to the value of the dictionary with key hereCanBeRandomNumber"
If you're trying to add to the dictionary, use:
jobs[hereCanBeRandomNumber] = job
This will add to the jobs dict, so it looks like:
jobs = {118096950: job}
So your problem here is you are trying to append to a key that does not exist yet.
Take a dictionary like
jobs = {}
and doing
jobs[123].append(foo)
will produce a KeyError as nothing exists at 123 yet.
To get around this you can do either of the following:
from collections import defaultdict
jobs = defaultdict(list)
jobs[123].append(foo)
which means if a key does not exist it is initialised to an empty list first or
jobs = {}
jobs[123] = jobs.get(123, []).append(job)
which checks jobs for the presence of the key and if it doesn't exist used an empty list
I have a list of dictionaries that maps different IDs to a central ID. I have a document with these different IDs associated with terms. I have created a function that now has a key the central ID from the different IDs in the document. The goFile is the document where in the first column there's an ID and in the second one there's a GOterm. The mappingList is a list containing dictionaries in which the ID in the goFile is mapped to a main ID.
My expected output is a dictionary with a main ID as a key and a set with the go terms associated with it as value.
def parseGO(mappingList, goFile):
# open the file
file = open(goFile)
# this will be the dictionary that this function returns
# entries will have as a key an Ensembl ID
# and the value will be a set of GO terms
GOdict = {}
GOset = set()
for line in file:
splitline = line.split(' ')
GO_term = splitline[1]
value_ID = splitline[0]
for dict in mappingList:
if value_ID in dict:
ENSB_term = dict[value_ID]
#my best try
for dict in mappingList:
for key in GOdict.keys():
if value_ID in dict and key == dict[value_ID]:
GOdict[ENSB_term].add(GO_term)
GOdict[ENSB_term] = GOset
return GOdict
My problem is that now I have to add to the central ID in my GOdict the terms that are associated in the document to the different IDs. To avoid duplicates i use a set (GOset). How do I do it? All my try end having all the terms mapped to all the main IDs.
Some sample:
mappingList = [{'1234': 'mainID1', '456': 'mainID2'}, {'789': 'mainID2'}]
goFile:
1234 GOTERM1
1234 GOTERM2
456 GOTERM1
456 GOTERM3
789 GOTERM1
expected output:
GOdict = {'mainID1': set([GOTERM1, GOTERM2]), 'mainID2': set([GOTERM1, GOTERM3])}
First off, you shouldn't use the variable name 'dict', as it shadows the built-in dict class, and will cause you problems at some point.
The following should work for you:
from collections import defaultdict
def parse_go(mapping_list, go_file):
go_dict = defaultdict(set)
with open(go_file) as f: # Better garbage handling using 'with'
for line in f:
(value_id, go_term) = line.split() # Feel free to change the split behaviour
# work better for you.
for map_dict in mapping_list:
if value_id in map_dict:
go_dict[map_dict[value_id]].add(go_term)
return go_dict
The code is fairly straightforward, but here's a breakdown anyway.
We use a default dictionary instead of a normal dictionary so we can eliminate all that if in or setdefault() boilerplate.
For each line in the file, we check if the first item (value_id) is a key in any of the mapping dictionaries, and if so, adds the lines second item (go_term) to that value_id's set in the dictionary.
EDIT: Request for doing this without defaultdict(). Assume that go_dict is just a normal dictionary (go_dict = {}), your for loop would look like:
for map_dict in mapping_list:
if value_id in map_dict:
esnb_entry = go_dict.setdefault(map_dict[value_id], set())
esnb_entry.add(go_term)