How to split a list into equally-sized chunks in Python?
The Good: Do include a small example DataFrame, either as runnable code: In [1]: df = pd.DataFrame([[1, 2], [1, 3], [4, 6]], columns=['A', 'B']) or make it "copy and pasteable" using pd.read_clipboard(sep=r'\s\s+'). In [2]: df Out[2]: A B 0 1 2 1 1 3 2 4 6 Test it yourself to make sure it works andRead more
The Good:
- Do include a small example DataFrame, either as runnable code:
In [1]: df = pd.DataFrame([[1, 2], [1, 3], [4, 6]], columns=['A', 'B'])
or make it “copy and pasteable” using
pd.read_clipboard(sep=r'\s\s+')
.In [2]: df Out[2]: A B 0 1 2 1 1 3 2 4 6
Test it yourself to make sure it works and reproduces the issue.
- You can format the text for Stack Overflow by highlighting and using Ctrl+K (or prepend four spaces to each line), or place three backticks (“`) above and below your code with your code unindented.
- I really do mean small. The vast majority of example DataFrames could be fewer than 6 rows,[citation needed] and I bet I can do it in 5. Can you reproduce the error with
df = df.head()
? If not, fiddle around to see if you can make up a small DataFrame which exhibits the issue you are facing.But every rule has an exception, the obvious one being for performance issues (in which case definitely use
%timeit
and possibly%prun
to profile your code), where you should generate:df = pd.DataFrame(np.random.randn(100000000, 10))
Consider using
np.random.seed
so we have the exact same frame. Having said that, “make this code fast for me” is not strictly on topic for the site. - For getting runnable code,
df.to_dict
is often useful, with the differentorient
options for different cases. In the example above, I could have grabbed the data and columns fromdf.to_dict('split')
.
- Write out the outcome you desire (similarly to above)
In [3]: iwantthis Out[3]: A B 0 1 5 1 4 6
Explain where the numbers come from:
The 5 is the sum of the B column for the rows where A is 1.
- Do show the code you’ve tried:
In [4]: df.groupby('A').sum() Out[4]: B A 1 5 4 6
But say what’s incorrect:
The A column is in the index rather than a column.
- Do show you’ve done some research (search the documentation, search Stack Overflow), and give a summary:
The docstring for sum simply states “Compute sum of group values”
The groupby documentation doesn’t give any examples for this.
Aside: the answer here is to use
df.groupby('A', as_index=False).sum()
. - If it’s relevant that you have Timestamp columns, e.g. you’re resampling or something, then be explicit and apply
pd.to_datetime
to them for good measure.df['date'] = pd.to_datetime(df['date']) # this column ought to be date.
Sometimes this is the issue itself: they were strings.
The Bad:
- Don’t include a MultiIndex, which we can’t copy and paste (see above). This is kind of a grievance with Pandas’ default display, but nonetheless annoying:
In [11]: df Out[11]: C A B 1 2 3 2 6
The correct way is to include an ordinary DataFrame with a
set_index
call:In [12]: df = pd.DataFrame([[1, 2, 3], [1, 2, 6]], columns=['A', 'B', 'C']) In [13]: df = df.set_index(['A', 'B']) In [14]: df Out[14]: C A B 1 2 3 2 6
- Do provide insight to what it is when giving the outcome you want:
B A 1 1 5 0
Be specific about how you got the numbers (what are they)… double check they’re correct.
- If your code throws an error, do include the entire stack trace. This can be edited out later if it’s too noisy. Show the line number and the corresponding line of your code which it’s raising against.
- Pandas 2.0 introduced a number of changes, and Pandas 1.0 before that, so if you’re getting unexpected output, include the version:
pd.__version__
On that note, you might also want to include the version of Python, your OS, and any other libraries. You could use
pd.show_versions()
or thesession_info
package (which shows loaded libraries and Jupyter/IPython environment).
The Ugly:
- Don’t link to a CSV file we don’t have access to (and ideally don’t link to an external source at all).
df = pd.read_csv('my_secret_file.csv') # ideally with lots of parsing options
Most data is proprietary, we get that. Make up similar data and see if you can reproduce the problem (something small).
- Don’t explain the situation vaguely in words, like you have a DataFrame which is “large”, mention some of the column names in passing (be sure not to mention their dtypes). Try and go into lots of detail about something which is completely meaningless without seeing the actual context. Presumably no one is even going to read to the end of this paragraph.
Essays are bad; it’s easier with small examples.
- Don’t include 10+ (100+??) lines of data munging before getting to your actual question.
Please, we see enough of this in our day jobs. We want to help, but not like this…. Cut the intro, and just show the relevant DataFrames (or small versions of them) in the step which is causing you trouble.
Here's a generator that yields evenly-sized chunks: def chunks(lst, n): """Yield successive n-sized chunks from lst.""" for i in range(0, len(lst), n): yield lst[i:i + n] import pprint pprint.pprint(list(chunks(range(10, 75), 10))) [[10, 11, 12, 13, 14, 15, 16, 17, 18, 19], [20, 21, 22, 23, 24, 25,Read more
Here’s a generator that yields evenly-sized chunks:
For Python 2, using
xrange
instead ofrange
:Below is a list comprehension one-liner. The method above is preferable, though, since using named functions makes code easier to understand. For Python 3:
For Python 2:
See less