Python Pandas GroupBy get list of groups

Question

I have a line of code:

g = x.groupby('Color')

The colors are Red, Blue, Green, Yellow, Purple, Orange, and Black. How do I return this list? For similar attributes, I use x.Attribute and it works fine, but x.Color doesn't behave the same way.

Answer 1

There is much easier way of doing it:

g = x.groupby('Color')

g.groups.keys()

By doing groupby() pandas returns you a dict of grouped DFs. You can easily get the key list of this dict by python built in function keys().

Answer 2

If you do not care about the order of the groups, Yanqi Ma's answer will work fine:

g = x.groupby('Color')
g.groups.keys()
list(g.groups) # or this

However, note that g.groups is a dictionary, so in Python <3.7 the keys are inherently unordered! This is the case even if you use sort=True on the groupby method to sort the groups, which is true by default.

This actually bit me hard when it resulted in a different order on two platforms, especially since I was using list(g.groups), so it wasn't obvious at first that g.groups was a dict.

In my opinion, the best way to do this is to take advantage of the fact that the GroupBy object has an iterator, and use a list comprehension to return the groups in the order they exist in the GroupBy object:

g = x.groupby('Color')
groups = [name for name,unused_df in g]

It's a little less readable, but this will always return the groups in the correct order.

Answer 3

Here's how to do it.

groups = list()
for g, data in x.groupby('Color'):
    print(g, data)
    groups.append(g)

The core idea here is this: if you iterate over a dataframe groupby iterator, you'll get back a two-tuple of (group name, filtered data frame), where filtered data frame contains only records corresponding to that group).

Answer 4

It is my understanding that you have a Data Frame which contains multiples columns. One of the columns is "Color" which has different types of colors. You want to return a list of unique colors that exist.

colorGroups = df.groupby(['Color'])
for c in colorGroups.groups: 
    print c

The above code will give you all the colors that exist without repeating the colors names. Thus, you should get an output such as:

Red
Blue
Green
Yellow
Purple
Orange
Black

An alternative is the unique() function which returns an array of all unique values in a Series. Thus to get an array of all unique colors, you would do:

df['Color'].unique()

The output is an array, so for example print df['Color'].unique()[3] would give you Yellow.

Answer 5

I compared runtime for the solutions above (with my data):

In [443]: d = df3.groupby("IND")

In [444]: %timeit groups = [name for name,unused_df in d]
377 ms ± 27.3 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [445]: % timeit  list(d.groups)
1.08 µs ± 47.3 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In [446]: % timeit d.groups.keys()
708 ns ± 7.18 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In [447]: % timeit df3['IND'].unique()
5.33 ms ± 128 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

it seems that the 'd.groups.keys()' is the best method.

Answer 6

Hope this helps.. Happy Coding :)

df = pd.DataFrame(data=[['red','1','1.5'],['blue','20','2.5'],['red','15','4']],columns=(['color','column1','column2']))

list_req = list(df.groupby('color').groups.keys())
print(list_req)