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copied!<p>Toy example. First make a <code>datetime</code> index. Here I make an index using two days repeated 10 times each. I then make some dummy data using <code>randn</code>.</p> <pre><code>In [1]: date_index = [datetime(2012,01,01)] * 10 + [datetime(2013,01,01)] * 10 In [2]: df = DataFrame({'A':randn(20),'B':randn(20)}, index=date_index) In [3]: df Out[3]: A B 2012-01-01 -1.155124 1.018059 2012-01-01 -0.312090 -1.083568 2012-01-01 0.688247 -1.296995 2012-01-01 -0.205218 0.837194 2012-01-01 0.700611 -0.001015 2012-01-01 1.996796 -0.914564 2012-01-01 -2.268237 0.517232 2012-01-01 -0.170778 -0.143245 2012-01-01 -0.826039 0.581035 2012-01-01 -0.351097 -0.013259 2013-01-01 -0.767911 -0.009232 2013-01-01 -0.322831 -1.384785 2013-01-01 0.300160 0.334018 2013-01-01 -1.406878 -2.275123 2013-01-01 1.722454 0.873262 2013-01-01 0.635711 -1.763352 2013-01-01 -0.816891 -0.451424 2013-01-01 -0.808629 -0.092290 2013-01-01 0.386046 -1.297096 2013-01-01 0.261837 0.562373 </code></pre> <p>If I understand your question correctly, you want to decile <em>within</em> each date. To do that, you can first move the index into the dataframe as a column. Then, you can groupby by the new column (here it's called index), and use <code>transform</code> with a lambda function. The lambda function below, applies <code>pandas.qcut</code> to the grouped <code>series</code> and returns the <code>labels</code> attribute.</p> <pre><code>In [4]: df.reset_index().groupby('index').transform(lambda x: qcut(x,10).labels) Out[4]: A B 0 1 9 1 4 1 2 7 0 3 5 8 4 8 5 5 9 2 6 0 6 7 6 3 8 2 7 9 3 4 10 3 6 11 4 2 12 6 7 13 0 0 14 9 9 15 8 1 16 1 4 17 2 5 18 7 3 19 5 8 </code></pre>

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