ValueError: Grouper for <something> not 1-dimensional
Simplified problem
I also ran into this problem and found that it was caused by duplicate column names.
To recreate this:
df = pd.DataFrame({"foo": [1,2,3], "bar": [1,2,3]})
df.rename(columns={'foo': 'bar'}, inplace=True)
bar bar
0 1 1
1 2 2
2 3 3
df.groupby('bar')
ValueError: Grouper for 'bar' not 1-dimensional
Just like a lot of cryptic pandas errors, this one too stems from having two columns with the same name.
Figure out which one you want to use, rename or drop the other column and redo the operation.
Solution
Rename the columns like this
df.columns = ['foo', 'bar']
foo bar
0 1 1
1 2 2
2 3 3
df.groupby('bar')
<pandas.core.groupby.DataFrameGroupBy object at 0x1066dd950>
TL;DR:
Quick example: if I am to groupby a bunch of people by careers, a person is either an eng or a tech, can't be both, otherwise groupby()
won't know if to put that person in the tech group or the eng group.
Your code, unfortunately assigned some people into both eng AND tech at the same time.
First of all, just to make sure we TRULY understand what groupby()
does.
We will be using this example fruit df
thru out:
import pandas as pd
import numpy as np
df = pd.DataFrame(
{"fruit": ['apple', 'apple', 'orange', 'orange'], "color": ['r', 'g', 'b', 'r']},
index=[11, 22, 33, 44],
)
"""
[df] df:
+----+---------+---------+
| | fruit | color |
|----+---------+---------|
| 11 | apple | r |
| 22 | apple | g |
| 33 | orange | b |
| 44 | orange | r |
+----+---------+---------+
"""
Below is a very valid df.groupby()
, not using any column names:
gp = df.groupby(
{
0: 'mine',
1: 'mine',
11: 'mine',
22: 'mine',
33: 'mine',
44: 'you are rats with wings!',
}
)
"""
[df] [group] mine:
+----+---------+---------+
| | fruit | color |
|----+---------+---------|
| 11 | apple | r |
| 22 | apple | g |
| 33 | orange | b |
+----+---------+---------+
[df] [group] you are rats with wings!:
+----+---------+---------+
| | fruit | color |
|----+---------+---------|
| 44 | orange | r |
+----+---------+---------+
"""
Wait, the groupby()
didn't even use 'fruit' or 'color' at all?!
That's right! groupby()
doesn't need to care about df
or 'fruit' or 'color' or Nemo, groupby()
only cares about one thing, a lookup table that tells it which index is mapped to which label (ie. group).
In this case, for example, the dictionary passed to the groupby()
is instructing the groupby()
to:
if you see index 11
, then it is a "mine"
, put the row with that index in the group named "mine"
.
if you see index 22
, then it is a "mine"
, put the row with that index in the group named "mine"
.
...
even 0 and 1 not being in df.index
is not a problem
Conventional df.groupby('fruit')
or df.groupby(df['fruit'])
follows exactly the rule above. The column df['fruit']
is used as a lookup table, it tells groupby()
that index 11
is an "apple"
Now, regarding: Grouper for '<class 'pandas.core.frame.DataFrame'>' not 1-dimensional
what it is saying is really: for some or all indexes in df, you are assigning MORE THAN just one label
Let's examine some possible errors using the above example:
[x] df.groupby(df)
will not work, you gave groupby()
a 2D mapping, each index was given 2 group names. It will complain: is index 11 an "apple" or an "r"? make up your mind!
[x] the below codes will also not work. Although the mapping is now 1D, it is mapping index 11
to "mine"
as well as "yours"
. Pandas' df
and sr
allow none-unique index, so be careful.
mapping = pd.DataFrame(index= [ 11, 11, 22, 33, 44 ],
data = ['mine', 'yours', 'mine', 'mine', 'yours'], )
df.groupby(mapping)
# different error message, but same idea
mapping = pd.Series( index= [ 11, 11, 22, 33, 44 ],
data = ['mine', 'yours', 'mine', 'mine', 'yours'], )
df.groupby(mapping)