Most efficient way to split up a dataframe into smaller dataframes

I am writing a python program that will parse a large dataframe (tens of thousands of lines) into smaller dataframes based on a column value, and it needs to be fairly efficient, because the user can change the ways they break up the dataframe, and I would like the output to update dynamically.

Example input:

Example Output:

The naïve way, in my mind, is to do something like this:

for id in list(df['id'].unique()):
    filtered_df = df[df['id'] == id]  

But I believe this would be looping over the same data more times than is necessary, which is inefficient. Is there a fast way of doing this?

Update

Did a little software drag racing. Here are the results:

%%timeit
[df.loc[df.id.eq(i)] for i in df.id.unique()]

9.96 ms ± 1.26 ms per loop (mean ± std. dev. of 7 runs, 100 loops each)

%%timeit
dflist=[]
dflist2=[]
for  k,v  in df.groupby(['id']):
    var='id'+str(k)
    dflist.append(var)
    globals()[var] = v
    dflist2.append(v)

1.28 ms ± 92.2 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)

%%timeit
d = {id:df[df.id==id] for id in df.id.unique()}

9.19 ms ± 885 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Does anyone know why the second solution would be so much faster?

>Solution :

here is one way to do it

# create lists to hold name and the trimmed data from df
dflist=[]
dflist2=[]
for  k,v  in df.groupby(['id']):
    var='id'+str(k)

# append dataframe name in a list
    dflist.append(var)

# create a DF with a variable name
    globals()[var] = v

# append dataframe in a list
    dflist2.append(v)

# to list all the dataframe created 
>>> dflist

['id1', 'id2', 'id3']

# accessing a dataframe
>>> id1


    id  Column_1    Column_2
0   1   Oct     10000$
1   1   Dec     9000$
5   1   Nov     15000$

>> dflist2

[   id Column_1 Column_2
 0   1     Oct    10000$
 1   1     Dec     9000$
 5   1     Nov    15000$,
    id Column_1 Column_2
 2   2     Oct     3400$
 4   2     Nov     9000$,
    id Column_1 Column_2
 3   3     Dec    20000$]