Pardon the noob question, I am new to Python.
I am working with a list of dictionaries:
dataset = [
{'id':1,'dateCollected':'2021-03-02','orders':7},
{'id':2,'dateCollected':'2021-03-03','orders':8},
]
.... this goes on for 50 records ....
I would like to a make a for loop that iterates through every dictionary in the list and adds certain key-value pairs to a new dictionary.
For Example:
match_history_data = {}
for i in dataset:
match_history_data['DateCollected'] = i['dateCollected']
match_history_data['Orders'] = i['orders']
print(match_history_data)
results in:
{'DateCollected': '2021-03-03', 'Orders': 8}
I would like the result to be:
{'DateCollected':'2021-03-02','Orders':7},
{'DateCollected':'2021-03-03','Orders':8}
This works how I want it to, but only for the first iteration in the for loop. How do I get around this so that it goes thru all the records in 'dataset', and creates the new list or dictionary?
You're creating a new dictionary, match_history_data and then just setting the same entries 'DateCollected' and 'Orders' over and over again. Each iteration of your for loop is just overwriting the same entries.
What you want is a list of new dictionaries:
match_history_data = []
for item in dataset:
match_history_data.append(
{"DateCollected" : item["dateCollected"],
"Orders" : item["orders"]})
You can achieve this in a neater fashion using a list comprehension:
match_history_data = [{"DateCollected" : item["dateCollected"], "Orders" : item["orders"]} for item in dataset]
How about this
match_history_data = [{"DateCollected":x["dateCollected"], "Orders":x["orders"]} for x in dataset]
What you have right now overwrites a single dictionary as your loop through the original list of dictionaries, since keys are unique in dictionaries. What you need to is another list to which you can append the new dictionaries during each iteration to a separate list.
For example
new_list = []
for i in dataset:
match_history_data = {}
match_history_data['DateCollected'] = i['dateCollected']
match_history_data['Orders'] = i['orders']
new_list.append(match_history_data)
If you don't want to preserve the original list, you can simply drop the id key in each dictionary.
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.
New to python and for this example list
lst = ['<name>bob</name>', '<job>doctor</job>', '<gender>male</gender>', '<name>susan</name>', '<job>teacher</job>', '<gender>female</gender>', '<name>john</name>', '<gender>male</gender>']
There are 3 categories of name, job, and gender. I would want those 3 categories to be on the same line which would look like
<name>bob</name>, <job>doctor</job>, <gender>male</gender>
My actual list is really big with 10 categories I would want to be on the same line. I am also trying to figure out a way where if one of the categories is not in the list, it would print something like N/A to indicate that it is not in the list
for example I would want it to look like
<name>bob</name>, <job>doctor</job>, <gender>male</gender>
<name>susan</name>, <job>teacher</job>, <gender>female</gender>
<name>john</name>, N/A, <gender>male</gender>
What would be the best way to do this?
This is one way to do it. This would handle any length list, and guarantee grouping no matter how long the lists are as long as they are in the correct order.
Updated to convert to dict, so you can test for key existence.
lst = ['<name>bob</name>', '<job>doctor</job>', '<gender>male</gender>', '<name>susan</name>', '<job>teacher</job>', '<gender>female</gender>', '<name>john</name>', '<gender>male</gender>']
newlst = []
tmplist = {}
for item in lst:
value = item.split('>')[1].split('<')[0]
key = item.split('<')[1].split('>')[0]
if '<name>' in item:
if tmplist:
newlst.append(tmplist)
tmplist = {}
tmplist[key] = value
#handle the remaining items left over in the list
if tmplist:
newlst.append(tmplist)
print(newlst)
#test for existance
for each in newlst:
print(each.get('job', 'N/A'))
i'm using an api call in python 3.7 which returns json data.
result = (someapicall)
the data returned appears to be in the form of two nested dictionaries within a list, i.e.
[{name:foo, firmware:boo}{name:foo, firmware:bar}]
i would like to retrieve the value of the key "name" from the first dictionary and also the value of key "firmware" from both dictionaries and store in a new dictionary in the following format.
{foo:(boo,bar)}
so far i've managed to retrieve the value of both the first "name" and the first "firmware" and store in a dictionary using the following.
dict1={}
for i in result:
dict1[(i["networkId"])] = (i['firmware'])
i've tried.
d7[(a["networkId"])] = (a['firmware'],(a['firmware']))
but as expected the above just seems to return the same firmware twice.
can anyone help achive the desired result above
you can use defaultdict to accumulate values in a list, like this:
from collections import defaultdict
result = [{'name':'foo', 'firmware':'boo'},{'name':'foo', 'firmware':'bar'}]
# create a dict with a default of empty list for non existing keys
dict1=defaultdict(list)
# iterate and add firmwares of same name to list
for i in result:
dict1[i['name']].append(i['firmware'])
# reformat to regular dict with tuples
final = {k:tuple(v) for k,v in dict1.items()}
print(final)
Output:
{'foo': ('boo', 'bar')}
I am trying to either
create dictionaries names based on loop index e.g. mydict_1, mydict_2 etc.
or
append dictionaries in one dictionary
Through a loop I am getting sets of data and I want to be able to access them all at once or one by one.
for components in fiSentence.findall("components"):
operation = components.find('operation').text
target = components.find('target').text
targetState = components.find('targetState').text
...
all this going in a dictionary:
tempDict = {"operation":operation, "target":target, "targetState":targetState, ...}
and then outside of the loop I tried to store all of them in another dictionary but I only managed to do so with a list:
data.append(tempDict)
What I want is either to store them in different dictionaries as:
procedural_Step_1 = {"operation":operation, "target":target, "targetState":targetState}
procedural_Step_2 = {"operation":operation, "target":target, "targetState":targetState}
...
or
store them all in one dictionary of dictionaries:
data = {"procedural_Step_1":{"operation":operation, "target":target, "targetState":targetState}, {"procedural_Step_2":{"operation":operation, "target":target, "targetState":targetState},...}
You can declare dict data before the loop and in the end of loop:
data['procedural_step_'+str(index)] = temp_dict
Index you can get with enumerate
I have a list of dictionaries called lod. All dictionaries have the same keys but different values. I am trying to update one specific value in the list of values for the same key in all the dictionaries.
I am attempting to do it with the following for loop:
for i in range(len(lod)):
a=lod[i][key][:]
a[p]=a[p]+lov[i]
lod[i][key]=a
What's happening is each is each dictionary is getting updated len(lod) times so lod[0][key][p] is supposed to have lov[0] added to it but instead it is getting lov[0]+lov[1]+.... added to it.
What am I doing wrong?
Here is how I declared the list of dicts:
lod = [{} for _ in range(len(dataul))]
for j in range(len(dataul)):
for i in datakl:
rrdict[str.split(i,',')[0]]=list(str.split(i,',')[1:len(str.split(i,','))])
lod[j]=rrdict
The problem is in how you created the list of dictionaries. You probably did something like this:
list_of_dicts = [{}] * 20
That's actually the same dict 20 times. Try doing something like this:
list_of_dicts = [{} for _ in range(20)]
Without seeing how you actually created it, this is only an example solution to an example problem.
To know for sure, print this:
[id(x) for x in list_of_dicts]
If you defined it in the * 20 method, the id is the same for each dict. In the list comprehension method, the id is unique.
This it where the trouble starts: lod[j] = rrdict. lod itself is created properly with different dictionaries. Unfortunately, afterwards any references to the original dictionaries in the list get overwritten with a reference to rrdict. So in the end, the list contains only references to one single dictionary. Here is some more pythonic and readable way to solve your problem:
lod = [{} for _ in range(len(dataul))]
for rrdict in lod:
for line in datakl:
splt = line.split(',')
rrdict[splt[0]] = splt[1:]
You created the list of dictionaries correctly, as per other answer.
However, when you are updating individual dictionaries, you completely overwrite the list.
Removing noise from your code snippet:
lod = [{} for _ in range(whatever)]
for j in range(whatever):
# rrdict = lod[j] # Uncomment this as a possible fix.
for i in range(whatever):
rrdict[somekey] = somevalue
lod[j] = rrdict
Assignment on the last line throws away the empty dict that was in lod[j] and inserts a reference to the object represented by rrdict.
Not sure what your code does, but see a commented-out line - it might be the fix you are looking for.