Let's say I have this loop with a definition in it:
for name in ["alice", "bob", "ceelo"]:
full_name = name + {"alice": "cooper", "bob": "dylan", "ceelo":"green"}[name]
print(full_name)
As you can see, my dict isn't assigned to anything. I could save runtime by refractoring it to:
names = {"alice": "cooper", "bob": "dylan", "ceelo":"green"}
for name in names:
full_name = name + names[name]
... but I don't want to. For reasons. I promise.
My question: Does the standard Python compiler automatically perform this refractor?
First, python does not use a compiler because there is no compiled file generated after you try to run some python files, it uses interpreter.
Here even if you do not assign the dict to any variable name e.g.
"names = {"alice": "cooper", "bob": "dylan", "ceelo":"green"}", the dictionary itself is still a dictionary and it occupies some space in your memory, no matter a name given or not.
Therefore
, you will be able to use key to get your value e.g. "{"alice": "cooper", "bob": "dylan", "ceelo":"green"}[name]". But the python interpreter will not assign the dictionary to a variable automatically if you never declare so.
Related
I'm trying to add a key to a dictionary with the update() method, but since the program might need to be closed, I need for the key to be saved backed into the file for later use.
If I add the key {'car' : 'purple'} to the dictionary it would save as:
dict = {
{'truck' : 'blue'},
{'mini' : 'green'},
{'car' : 'purple'}
}
instead of in its previous value.
dict = {
{'truck' : 'blue'},
{'mini' : 'green'}
}
When you assign a value to a variable, for example the dict type you show, it is not saved in any file. If you want to persist the variable to a file you will need to do something special.
If you work in the Python shell (aka REPL) you can use shelve to persist the name/value pairs.
If you want your own program to save data, that data could be written to a JSON file, or saved in a database.
BTW, it is considered bad practice to name a variable dict because that is also a type.
Also, your use of the term default state would be better expressed as previous value, because you have not indicated that this variable is assigned from a property in an API or framework.
You can save the file simply to a text file and import it when need be.
dict = {'truck': 'blue', 'mini': 'green', 'car': 'purple'}
new_file = open('newfile.txt', 'wt')
new_file.write(str(dict))
new_file.close()
I'm trying to create a function that returns a dynamically-named list of columns. Usually I can manually name the list, but I now have 100+ csv files to work with.
My goal:
Function creates a list, and names it based on dataframe name
Created list is callable outside of the function
I've done my research, and this answer from an earlier post came very close to helping me.
Here is what I've adapted
def test1(dataframe):
# Using globals() to get dataframe name
df_name = [x for x in globals() if globals()[x] is dataframe][0]
# Creating local dictionary to use exec function
local_dict = {}
# Trying to generate a name for the list, based on input dataframe name
name = 'col_list_' + df_name
exec(name + "=[]", globals(), local_dict)
# So I can call this list outside the function
name = local_dict[name]
for feature in dataframe.columns:
# Append feature/column if >90% of values are missing
if dataframe[feature].isnull().mean() >= 0.9:
name.append(feature)
return name
To ensure the list name changes based on the DataFrame supplied to the function, I named the list using:
name = 'col_list_' + df_name
The problem comes when I try to make this list accessible outside the function:
name = local_dict[name].
I cannot find away to assign a dynamic list name to the local dictionary, so I am forced to always call name outside the function to return the list. I want the list to be named based on the dataframe input (eg. col_list_df1, col_list_df2, col_list_df99).
This answer was very helpful, but it seems specific to variables.
global 'col_list_' + df_name returns a syntax error.
Any help would be greatly appreciated!
I have read almost every similar question but none of them seems to solve my current problem.
In my python code, I am importing a string from my bashrc and in the following, I am defining the same name as a variable to index my dictionary. Here is the simple example
obs = os.environ['obs']
>> obs = 'id_0123'
id_0123 = numpy.where(position1 == 456)
>> position1[id_0123] = 456
>> position2[id_0123] = 789
But of course, when I do positions[obs], it throws an error since it is a string rather than an index (numpy.int64). So I have tried to look for a solution to convert my string into a variable but all solution suggesting to either convert into a dictionary or something else and assign the string to an integer, But I can not do that since my string is dynamic and will constantly change. In the end, I am going to have about 50 variables and I need to check the current obs corresponding to which variable, so I could use it as indices to access the parameters.
Edit:
Position1 and Position 2 are just bumpy arrays, so depending on the output of os.environ (which is 'id_0123' in this particular case), they will print an array element. So I can not assign 'id_0123' another string or number since I am using that exact name as a variable.
The logic is that there are many different arrays, I want to use the output of os.environ as an input to access the element of these arrays.
If you wanted to use a dictionary instead, this would work.
obs = 'id_0123'
my_dict = {obs: 'Tester'}
print (my_dict [obs])
print (my_dict ['id_0123'])
You could use the fact that (almost) everything is a dictionary in Python to create storage container that you index with obs:
container = lambda:None
container.id_0123 = ...
...
print(positions[container.__dict__[obs]])
Alternatively, you can use globals() or locals() to achieve the desired behavior:
import numpy
import os
def environment_to_variable(environment_key='obs', variable_values=globals()):
# Get the name of the variable we want to reference from the bash environment
variable_name = os.environ[environment_key]
# Grab the variable value that matches the named variable from the environment. Variable you want to reference must be visible in the global namespace by default.
variable_value = variable_values[variable_name]
return variable_value
positions = [2, 3, 5, 7, 11, 13]
id_0123 = 2 # could use numpy.where here
id_0456 = 4
index = environment_to_variable(variable_values=locals())
print(positions[index])
If you place this in a file called a.py, you can observe the following:
$ obs="id_0123" python ./a.py
5
$ obs="id_0456" python ./a.py
11
This allows you to index the array differently at runtime, which is what it seems like your intention is.
I have a series of dictionaries which each contain the same keys but their values are different i.e Age in dictionary 1 = 2, Age in dictionary 2 = 4 etc etc but they are broadly identical in structure.
what I would like to do is to randomly select one of these dictionaries and then assign specific values with the dictionary to variables. i.e python randomly chooses Dictionary 1 and then I then want to fill the dictAge variable with the age value from Dictionary 1.
import random
dictList = ['myDict', 'otherDict']
mydict = {
'age' : 10,
'other': "dummy data"
}
.
.
.
randomDict = random.choice(dictList)
dictAge = randomDict['age']
print(dictAge)
In the case of the code above what should happen is:
randomDict is assigned a random value from the distList variable (at the top). This sets which dictionary's values will be used going forward.
I next want the dictAge variable to then be assigned the age value from the selected dictionary. In this case (as mydict is was the only dictionary available) it should be assigned the age value of 10.
The error I am getting is:
TypeError: string indices must be integers
I know this is such a common error but my brain can't quite work out what the best solution is.
(Disclaimer: I haven't used python in ages so I know I am doing something really obviously silly but I can't quite work out what to do).
Right now, you are not actually using the definition of your dicts.
This is because dictList is comprised of strings: ['myDict', 'otherDict'].
So, when doing randomDict = random.choice(dictList), randomDict will either be the string 'myDict', or the string 'otherDict'.
Then you are doing randomDict['age'], which means you are trying to slice a string, with a string. As the error suggests, this can't be done and indices can only be ints.
What you want to do, is move the definition of the dictList to be after the definitions of your dicts, and include references to the dicts themselves, not strings. Something like:
mydict = {
'age' : 10,
'other': "dummy data"
}
.
.
.
dictList = [myDict, otherDict]
In the following piece of code:
dictAge = randomDict['age']
You are trying to index the name of dictionary variable (a string) returned by random.choice function.
To make it work you would need to do it using locals:
locals()[randomDict]['age']
or rather correct the dictList to contain the dictionaries instead of their names:
dictList = [myDict, otherDict]
In the latter case please note that myDict and otherDict should be declared before dictList.
I have a list of variables I need to run though a function to produce a json table. I want to loop through this list (list_db) and create a new variable to look through them manually in spyder. I am having trouble creating those new variables from the for loop. I thought i could use the name of the items in list as the new variable name, but i cant get it to work. Here is what I have:
for i in list_db:
p = str(i)
p = getDF(i) #function to run
What am I missing? What is the more standard way of doing this i cant think of?
It seems variable names don't really act how you are expecting. When you do p = str(i), you are assigning to the variable p. And then your next line, p = getDF(i) overwrites the value of p. It does not write to a variable whose name is str(i) like you are expecting.
If you want to store values into named slots, you can use a dictionary:
p = {}
for i in list_db:
p[i] = getDF(i)