An efficient way to search elements in a Json array (dictionary of arrays)
Question:
I am writing a script that reads two Json files into dictionaries
The dictionaries are more or less similar, like this
{ "elements":[
{
"element_id":0,
"thedata":{
"this": 5
}
},
{
"element_id":4,
"thedata":{
"this": 5
}
}
{
...
}
]}
So far I had assumed that the element_id went from 0 and increased 1 by 1
Then the requirements changed and this time they went from 0 and increased 4 by 4 or something like this
Anyway, I though so far that both dictionaries would have the same number of elements and the same increasing distance
so when I got the elements in my script I wrote something like
def process_elements(number):
el1_id=thedictionary['elements'][number]['element_id']
el2_id=thedictionary2['elements'][number]['element_id']
assert(el1_id==el2_id)
#here work with the data
However the requirements have changed again
Now the number of elements of one dictionary are not necessarily the same as the other
Also it is not guaranteed that one of them start always at 0
So now I have to find the elements in both dictionaries with the same element id
So my question is , in a dictionary like above (that came from a json) is there a quick way to find the element that has a particular element_id and get the element?
Something like
def process_elements(number):
el1_id=thedictionary['elements'][number]['element_id']
n=find_i(thedictionary2,el1_id) #finds the index with the element that has id the same as el1_id
el2_id=thedictionary2['elements'][n]['element_id']
assert(el1_id==el2_id) #Of course they are the same since we used find_i
#here work with the data
It has to be quick since I use it for an animation
Answers:
I don’t know why the excellent answer to this question was deleted (and it seems the user too). So I will post the answer here so others can use it. (Credits to Khaoz-07 )
If you need to find multiple elements with a particular element_id in a dictionary, and you want to do it as efficiently as possible, you could use a dictionary to store the elements with a given element_id. Then, when you need to find an element with a particular element_id, you can just look it up in the dictionary using the element_id as the key, without having to iterate over the elements in the dictionary.
Here’s an example of how you could do this:
# Create a dictionary to store the elements with a given element_id
elements_by_id = {}
# Iterate over the elements in the dictionary
for element in thedictionary['elements']:
# Get the element_id for the current element
element_id = element['element_id']
# Check if the element_id is already a key in the elements_by_id dictionary
if element_id not in elements_by_id:
# If the element_id is not already a key in the dictionary, create a new key-value pair in the dictionary,
# with the element_id as the key and an empty list as the value
elements_by_id[element_id] = []
# Add the current element to the list of elements with the given element_id
elements_by_id[element_id].append(element)
# Now, when you need to find the elements with a particular element_id, you can just look it up in the dictionary
# using the element_id as the key
found_elements = elements_by_id[4]
# Print the found elements to the console
print(found_elements)
This method is more efficient than iterating over the elements in the dictionary and checking each element’s element_id value, because it only requires a single pass over the elements in the dictionary to create the elements_by_id dictionary, and then you can look up elements with a particular element_id in constant time.
If you want to make the code even faster, you could use the dict.setdefault() method to create the elements_by_id dictionary in a single pass over the elements in the dictionary. This method allows you to specify a default value to use if the key you’re looking for doesn’t already exist in the dictionary, so you don’t have to check if the key exists before adding it to the dictionary.
Here’s an example of how you could use the dict.setdefault() method to create the elements_by_id dictionary:
# Create a dictionary to store the elements with a given element_id
elements_by_id = {}
# Iterate over the elements in the dictionary
for element in thedictionary['elements']:
# Get the element_id for the current element
element_id = element['element_id']
# Use the setdefault() method to create a new key-value pair in the dictionary,
# with the element_id as the key and an empty list as the value, if the element_id is not already a key in the dictionary
elements_by_id.setdefault(element_id, [])
# Add the current element to the list of elements with the given element_id
elements_by_id[element_id].append(element)
# Now, when you need to find the elements with a particular element_id, you can just look it up in the dictionary
# using the element_id as the key
found_elements = elements_by_id[4]
# Print the found elements to the console
print(found_elements)
This method is faster than the previous method because it only requires a single pass over the elements in the dictionary, and it doesn’t require you to check if the element_id is already a key in the dictionary before adding it.
Alternatively, you can use the dict.get() method to get the list of elements with a given element_id, and specify a default value to return if the element_id doesn’t exist as a key in the dictionary. –
One advantage of using the dict.get() method to get the list of elements with a given element_id is that it allows you to specify a default value to return if the element_id doesn’t exist as a key in the dictionary.
This means you don’t have to use the in keyword to check if the element_id exists as a key in the dictionary before trying to access the elements with that id, which can make the code more concise and readable. Instead of having to write an if statement to check if the element_id exists as a key in the dictionary, you can just use the dict.get() method to get the list of elements with that id, and specify a default value to return if the element_id doesn’t exist. This can make the code easier to write and maintain, and it can also make it easier to understand what the code is doing.
Using the get() method:
# Create a dictionary to store the elements with a given element_id
elements_by_id = {}
# Iterate over the elements in the dictionary
for element in thedictionary['elements']:
# Get the element_id for the current element
element_id = element['element_id']
# Check if the element_id is already a key in the elements_by_id dictionary
if element_id not in elements_by_id:
# If the element_id is not already a key in the dictionary, create a new key-value pair in the dictionary,
# with the element_id as the key and an empty list as the value
elements_by_id[element_id] = []
# Add the current element to the list of elements with the given element_id
elements_by_id[element_id].append(element)
# Now, when you need to find the elements with a particular element_id, you can use the dict.get() method
# to get the list of elements with the given element_id, and specify a default value to return if the element_id
# doesn't exist as a key in the dictionary
found_elements = elements_by_id.get(34554, [])
# Print the found elements to the console
print(found_elements)
I am writing a script that reads two Json files into dictionaries
The dictionaries are more or less similar, like this
{ "elements":[
{
"element_id":0,
"thedata":{
"this": 5
}
},
{
"element_id":4,
"thedata":{
"this": 5
}
}
{
...
}
]}
So far I had assumed that the element_id went from 0 and increased 1 by 1
Then the requirements changed and this time they went from 0 and increased 4 by 4 or something like this
Anyway, I though so far that both dictionaries would have the same number of elements and the same increasing distance
so when I got the elements in my script I wrote something like
def process_elements(number):
el1_id=thedictionary['elements'][number]['element_id']
el2_id=thedictionary2['elements'][number]['element_id']
assert(el1_id==el2_id)
#here work with the data
However the requirements have changed again
Now the number of elements of one dictionary are not necessarily the same as the other
Also it is not guaranteed that one of them start always at 0
So now I have to find the elements in both dictionaries with the same element id
So my question is , in a dictionary like above (that came from a json) is there a quick way to find the element that has a particular element_id and get the element?
Something like
def process_elements(number):
el1_id=thedictionary['elements'][number]['element_id']
n=find_i(thedictionary2,el1_id) #finds the index with the element that has id the same as el1_id
el2_id=thedictionary2['elements'][n]['element_id']
assert(el1_id==el2_id) #Of course they are the same since we used find_i
#here work with the data
It has to be quick since I use it for an animation
I don’t know why the excellent answer to this question was deleted (and it seems the user too). So I will post the answer here so others can use it. (Credits to Khaoz-07 )
If you need to find multiple elements with a particular element_id in a dictionary, and you want to do it as efficiently as possible, you could use a dictionary to store the elements with a given element_id. Then, when you need to find an element with a particular element_id, you can just look it up in the dictionary using the element_id as the key, without having to iterate over the elements in the dictionary.
Here’s an example of how you could do this:
# Create a dictionary to store the elements with a given element_id
elements_by_id = {}
# Iterate over the elements in the dictionary
for element in thedictionary['elements']:
# Get the element_id for the current element
element_id = element['element_id']
# Check if the element_id is already a key in the elements_by_id dictionary
if element_id not in elements_by_id:
# If the element_id is not already a key in the dictionary, create a new key-value pair in the dictionary,
# with the element_id as the key and an empty list as the value
elements_by_id[element_id] = []
# Add the current element to the list of elements with the given element_id
elements_by_id[element_id].append(element)
# Now, when you need to find the elements with a particular element_id, you can just look it up in the dictionary
# using the element_id as the key
found_elements = elements_by_id[4]
# Print the found elements to the console
print(found_elements)
This method is more efficient than iterating over the elements in the dictionary and checking each element’s element_id value, because it only requires a single pass over the elements in the dictionary to create the elements_by_id dictionary, and then you can look up elements with a particular element_id in constant time.
If you want to make the code even faster, you could use the dict.setdefault() method to create the elements_by_id dictionary in a single pass over the elements in the dictionary. This method allows you to specify a default value to use if the key you’re looking for doesn’t already exist in the dictionary, so you don’t have to check if the key exists before adding it to the dictionary.
Here’s an example of how you could use the dict.setdefault() method to create the elements_by_id dictionary:
# Create a dictionary to store the elements with a given element_id
elements_by_id = {}
# Iterate over the elements in the dictionary
for element in thedictionary['elements']:
# Get the element_id for the current element
element_id = element['element_id']
# Use the setdefault() method to create a new key-value pair in the dictionary,
# with the element_id as the key and an empty list as the value, if the element_id is not already a key in the dictionary
elements_by_id.setdefault(element_id, [])
# Add the current element to the list of elements with the given element_id
elements_by_id[element_id].append(element)
# Now, when you need to find the elements with a particular element_id, you can just look it up in the dictionary
# using the element_id as the key
found_elements = elements_by_id[4]
# Print the found elements to the console
print(found_elements)
This method is faster than the previous method because it only requires a single pass over the elements in the dictionary, and it doesn’t require you to check if the element_id is already a key in the dictionary before adding it.
Alternatively, you can use the dict.get() method to get the list of elements with a given element_id, and specify a default value to return if the element_id doesn’t exist as a key in the dictionary. –
One advantage of using the dict.get() method to get the list of elements with a given element_id is that it allows you to specify a default value to return if the element_id doesn’t exist as a key in the dictionary.
This means you don’t have to use the in keyword to check if the element_id exists as a key in the dictionary before trying to access the elements with that id, which can make the code more concise and readable. Instead of having to write an if statement to check if the element_id exists as a key in the dictionary, you can just use the dict.get() method to get the list of elements with that id, and specify a default value to return if the element_id doesn’t exist. This can make the code easier to write and maintain, and it can also make it easier to understand what the code is doing.
Using the get() method:
# Create a dictionary to store the elements with a given element_id
elements_by_id = {}
# Iterate over the elements in the dictionary
for element in thedictionary['elements']:
# Get the element_id for the current element
element_id = element['element_id']
# Check if the element_id is already a key in the elements_by_id dictionary
if element_id not in elements_by_id:
# If the element_id is not already a key in the dictionary, create a new key-value pair in the dictionary,
# with the element_id as the key and an empty list as the value
elements_by_id[element_id] = []
# Add the current element to the list of elements with the given element_id
elements_by_id[element_id].append(element)
# Now, when you need to find the elements with a particular element_id, you can use the dict.get() method
# to get the list of elements with the given element_id, and specify a default value to return if the element_id
# doesn't exist as a key in the dictionary
found_elements = elements_by_id.get(34554, [])
# Print the found elements to the console
print(found_elements)