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Asked: 2026-06-11T08:53:17+00:00

I have the following data frame in IPython, where each row is a single

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I have the following data frame in IPython, where each row is a single stock:

In [261]: bdata
Out[261]:
<class 'pandas.core.frame.DataFrame'>
Int64Index: 21210 entries, 0 to 21209
Data columns:
BloombergTicker      21206  non-null values
Company              21210  non-null values
Country              21210  non-null values
MarketCap            21210  non-null values
PriceReturn          21210  non-null values
SEDOL                21210  non-null values
yearmonth            21210  non-null values
dtypes: float64(2), int64(1), object(4)

I want to apply a groupby operation that computes cap-weighted average return across everything, per each date in the “yearmonth” column.

This works as expected:

In [262]: bdata.groupby("yearmonth").apply(lambda x: (x["PriceReturn"]*x["MarketCap"]/x["MarketCap"].sum()).sum())
Out[262]:
yearmonth
201204      -0.109444
201205      -0.290546

But then I want to sort of “broadcast” these values back to the indices in the original data frame, and save them as constant columns where the dates match.

In [263]: dateGrps = bdata.groupby("yearmonth")

In [264]: dateGrps["MarketReturn"] = dateGrps.apply(lambda x: (x["PriceReturn"]*x["MarketCap"]/x["MarketCap"].sum()).sum())
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
/mnt/bos-devrnd04/usr6/home/espears/ws/Research/Projects/python-util/src/util/<ipython-input-264-4a68c8782426> in <module>()
----> 1 dateGrps["MarketReturn"] = dateGrps.apply(lambda x: (x["PriceReturn"]*x["MarketCap"]/x["MarketCap"].sum()).sum())

TypeError: 'DataFrameGroupBy' object does not support item assignment

I realize this naive assignment should not work. But what is the “right” Pandas idiom for assigning the result of a groupby operation into a new column on the parent dataframe?

In the end, I want a column called “MarketReturn” than will be a repeated constant value for all indices that have matching date with the output of the groupby operation.

One hack to achieve this would be the following:

marketRetsByDate  = dateGrps.apply(lambda x: (x["PriceReturn"]*x["MarketCap"]/x["MarketCap"].sum()).sum())

bdata["MarketReturn"] = np.repeat(np.NaN, len(bdata))

for elem in marketRetsByDate.index.values:
    bdata["MarketReturn"][bdata["yearmonth"]==elem] = marketRetsByDate.ix[elem]

But this is slow, bad, and unPythonic.

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1 Answer

  1. Editorial Team
    In [97]: df = pandas.DataFrame({'month': np.random.randint(0,11, 100), 'A': np.random.randn(100), 'B': np.random.randn(100)})

In [98]: df.join(df.groupby('month')['A'].sum(), on='month', rsuffix='_r')
Out[98]:
           A         B  month       A_r
0  -0.040710  0.182269      0 -0.331816
1  -0.004867  0.642243      1  2.448232
2  -0.162191  0.442338      4  2.045909
3  -0.979875  1.367018      5 -2.736399
4  -1.126198  0.338946      5 -2.736399
5  -0.992209 -1.343258      1  2.448232
6  -1.450310  0.021290      0 -0.331816
7  -0.675345 -1.359915      9  2.722156
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