Pandas groupby to to_csv

50,871

Solution 1

Try doing this:

week_grouped = df.groupby('week')
week_grouped.sum().reset_index().to_csv('week_grouped.csv')

That'll write the entire dataframe to the file. If you only want those two columns then,

week_grouped = df.groupby('week')
week_grouped.sum().reset_index()[['week', 'count']].to_csv('week_grouped.csv')

Here's a line by line explanation of the original code:

# This creates a "groupby" object (not a dataframe object) 
# and you store it in the week_grouped variable.
week_grouped = df.groupby('week')

# This instructs pandas to sum up all the numeric type columns in each 
# group. This returns a dataframe where each row is the sum of the 
# group's numeric columns. You're not storing this dataframe in your 
# example.
week_grouped.sum() 

# Here you're calling the to_csv method on a groupby object... but
# that object type doesn't have that method. Dataframes have that method. 
# So we should store the previous line's result (a dataframe) into a variable 
# and then call its to_csv method.
week_grouped.to_csv('week_grouped.csv')

# Like this:
summed_weeks = week_grouped.sum()
summed_weeks.to_csv('...')

# Or with less typing simply
week_grouped.sum().to_csv('...')

Solution 2

Try changing your second line to week_grouped = week_grouped.sum() and re-running all three lines.

If you run week_grouped.sum() in its own Jupyter notebook cell, you'll see how the statement returns the output to the cell's output, instead of assigning the result back to week_grouped. Some pandas methods have an inplace=True argument (e.g., df.sort_values(by=col_name, inplace=True)), but sum does not.

EDIT: does each week number only appear once in your CSV? If so, here's a simpler solution that doesn't use groupby:

df = pd.read_csv('input.csv')
df[['id', 'count']].to_csv('output.csv')

Solution 3

Group By returns key, value pairs where key is the identifier of the group and the value is the group itself, i.e. a subset of an original df that matched the key.

In your example week_grouped = df.groupby('week') is set of groups (pandas.core.groupby.DataFrameGroupBy object) which you can explore in detail as follows:

for k, gr in week_grouped:
    # do your stuff instead of print
    print(k)
    print(type(gr)) # This will output <class 'pandas.core.frame.DataFrame'>
    print(gr)
    # You can save each 'gr' in a csv as follows
    gr.to_csv('{}.csv'.format(k))

Or alternatively you can compute aggregation function on your grouped object

result = week_grouped.sum()
# This will be already one row per key and its aggregation result
result.to_csv('result.csv') 

In your example you need to assign the function result to some variable as by default pandas objects are immutable.

some_variable = week_grouped.sum() 
some_variable.to_csv('week_grouped.csv') # This will work

basically result.csv and week_grouped.csv are meant to be same

Solution 4

I feel that there is no need to use a groupby, you can just drop the columns you do not want too.

df = df.drop(['month','year'], axis=1)
df.reset_index()
df.to_csv('Your path')

Solution 5

Pandas groupby generates a lot of information (count, mean, std, ...). If you want to save all of them in a csv file, first you need to convert it to a regular Dataframe:

import pandas as pd
...
...
MyGroupDataFrame = MyDataFrame.groupby('id')
pd.DataFrame(MyGroupDataFrame.describe()).to_csv("myTSVFile.tsv", sep='\t', encoding='utf-8')
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50,871
kalmdown
Author by

kalmdown

Updated on September 25, 2021

Comments

  • kalmdown
    kalmdown over 2 years

    Want to output a Pandas groupby dataframe to CSV. Tried various StackOverflow solutions but they have not worked.

    Python 3.6.1, Pandas 0.20.1

    groupby result looks like:

    id  month   year    count
    week                
    0   9066    82  32142   895
    1   7679    84  30112   749
    2   8368    126 42187   872
    3   11038   102 34165   976
    4   8815    117 34122   767
    5   10979   163 50225   1252
    6   8726    142 38159   996
    7   5568    63  26143   582
    

    Want a csv that looks like

    week  count
    0   895
    1   749
    2   872
    3   976
    4   767
    5   1252
    6   996
    7   582
    

    Current code:

    week_grouped = df.groupby('week')
    week_grouped.sum() #At this point you have the groupby result
    week_grouped.to_csv('week_grouped.csv') #Can't do this - .to_csv is not a df function. 
    

    Read SO solutions:

    output groupby to csv file pandas

    week_grouped.drop_duplicates().to_csv('week_grouped.csv')
    

    Result: AttributeError: Cannot access callable attribute 'drop_duplicates' of 'DataFrameGroupBy' objects, try using the 'apply' method

    Python pandas - writing groupby output to file

    week_grouped.reset_index().to_csv('week_grouped.csv')
    

    Result: AttributeError: "Cannot access callable attribute 'reset_index' of 'DataFrameGroupBy' objects, try using the 'apply' method"

    • cs95
      cs95 about 5 years
      If you landed up here wanting to know how to save each individual groupby to its own CSV file, see this answer.
  • kalmdown
    kalmdown over 6 years
    Thanks! - Why does it work when sum() is part of the the to_csv statement but not when sum() is done on its own line?
  • kalmdown
    kalmdown over 6 years
    In the original data the week appears on multiple rows. In this case the groupby is being used to gather the weeks together so a count can be done per week.
  • kalmdown
    kalmdown over 6 years
    BTW - Thanks so much for the explanation of why sum is an issue.
  • kalmdown
    kalmdown over 6 years
    Thank you for the indepth explanation. Helps to understand the system instead of just the problem.
  • kalmdown
    kalmdown over 6 years
    Should be "axis=1"...But yes that will output the rows but not grouped by week or state.
  • Alex Luis Arias
    Alex Luis Arias about 5 years
    @kalmdown, if this answered your question, can you please mark it as so? Click the check mark to make it green.
  • Alex Luis Arias
    Alex Luis Arias almost 4 years
    @kalmdown, did my reply answer your question? My answer still hasn't been marked as accepted.