Release 1.1.1 License BSD 3-Clause License Copyright (c) 2008-2011, AQR Capital Management, LLC, Lambda Foundry, Inc. and PyData ˓→Development Team All rights reserved. Copyright (c) 2011-2020, Open 268733 0.965327 1.685864 25 10.0 2013-01-06 0.778233 0.218139 2.659708 30 15.0 In [67]: df.apply(lambda x: x.max() - x.min()) Out[67]: A 1.787337 B 1.935455 C 1.953471 D 0.000000 F 4.000000 dtype: one argument to be evaluated on the DataFrame being assigned to. In [80]: iris.assign(sepal_ratio=lambda x: (x['SepalWidth'] / x['SepalLength'])). ˓→head() Out[80]: SepalLength SepalWidth PetalLength

Release 1.1.0 License BSD 3-Clause License Copyright (c) 2008-2011, AQR Capital Management, LLC, Lambda Foundry, Inc. and PyData ˓→Development Team All rights reserved. Copyright (c) 2011-2020, Open 985543 1.475647 -1.032412 25 10.0 2013-01-06 3.817014 1.246835 0.085539 30 15.0 In [67]: df.apply(lambda x: x.max() - x.min()) Out[67]: A 1.663374 B 3.119199 C 2.164186 D 0.000000 F 4.000000 dtype: one argument to be evaluated on the DataFrame being assigned to. In [80]: iris.assign(sepal_ratio=lambda x: (x['SepalWidth'] / x['SepalLength'])). ˓→head() Out[80]: SepalLength SepalWidth PetalLength

function for broadcasting results. In [7]: df.transform(['abs', lambda x: x - x.min()]) Out[7]: A B C abs <lambda> abs <lambda> abs <lambda> 2000-01-01 1.474071 2.671143 0.064034 1.007322 1.282782 0.000000 089354 9 10.0 -0.129820 0.631523 -0.586538 0.290720 In [58]: styled = df.style.\ ....: applymap(lambda val: 'color: %s' % 'red' if val < 0 else 'black').\ ....: highlight_max() ....: In [59]: styled Behavior: In [5]: idx.map(lambda x: x * 2) Out[5]: array([2, 4]) In [6]: idx.map(lambda x: (x, x * 2)) Out[6]: array([(1, 2), (2, 4)], dtype=object) In [7]: mi.map(lambda x: x) Out[7]: array([(1, 2)

function for broadcasting results. In [7]: df.transform(['abs', lambda x: x - x.min()]) Out[7]: A B C abs <lambda> abs <lambda> abs <lambda> 2000-01-01 1.474071 2.671143 0.064034 1.007322 1.282782 0.000000 089354 9 10.0 -0.129820 0.631523 -0.586538 0.290720 In [58]: styled = df.style.\ ....: applymap(lambda val: 'color: %s' % 'red' if val < 0 else 'black').\ ....: highlight_max() ....: In [59]: styled 4]], labels=[[0, 1], [0, 1]]) Previous Behavior: In [5]: idx.map(lambda x: x * 2) Out[5]: array([2, 4]) In [6]: idx.map(lambda x: (x, x * 2)) 1.2. v0.20.1 (May 5, 2017) 21 pandas: powerful Python

website page. License BSD 3-Clause License Copyright (c) 2008-2011, AQR Capital Management, LLC, Lambda Foundry, Inc. and PyData ˓→Development Team All rights reserved. Copyright (c) 2011-2021, Open -2.417535 -2.648122 25 10.0 2013-01-06 -0.026844 -2.303886 -4.126549 30 15.0 In [67]: df.apply(lambda x: x.max() - x.min()) Out[67]: A 2.073961 B 2.671590 C 1.785291 D 0.000000 F 4.000000 dtype: one argument to be evaluated on the DataFrame being assigned to. In [85]: iris.assign(sepal_ratio=lambda x: (x["SepalWidth"] / x["SepalLength"])). ˓→head() Out[85]: SepalLength SepalWidth PetalLength

website page. License BSD 3-Clause License Copyright (c) 2008-2011, AQR Capital Management, LLC, Lambda Foundry, Inc. and PyData␣ ˓→Development Team All rights reserved. Copyright (c) 2011-2021, Open -2.417535 -2.648122 25 10.0 2013-01-06 -0.026844 -2.303886 -4.126549 30 15.0 In [67]: df.apply(lambda x: x.max() - x.min()) Out[67]: A 2.073961 B 2.671590 C 1.785291 D 0.000000 F 4.000000 dtype: 187 pandas: powerful Python data analysis toolkit, Release 1.3.3 In [85]: iris.assign(sepal_ratio=lambda x: (x["SepalWidth"] / x["SepalLength"])).head() Out[85]: SepalLength SepalWidth PetalLength PetalWidth

website page. License BSD 3-Clause License Copyright (c) 2008-2011, AQR Capital Management, LLC, Lambda Foundry, Inc. and PyData␣ ˓→Development Team All rights reserved. Copyright (c) 2011-2021, Open -2.417535 -2.648122 25 10.0 2013-01-06 -0.026844 -2.303886 -4.126549 30 15.0 In [67]: df.apply(lambda x: x.max() - x.min()) Out[67]: A 2.073961 B 2.671590 C 1.785291 D 0.000000 F 4.000000 dtype: 187 pandas: powerful Python data analysis toolkit, Release 1.3.4 In [85]: iris.assign(sepal_ratio=lambda x: (x["SepalWidth"] / x["SepalLength"])).head() Out[85]: SepalLength SepalWidth PetalLength PetalWidth

35 3 35 36 3 36 37 3 37 38 3 38 39 3 39 [40 rows x 2 columns] In [9]: df.groupby('A').apply(lambda x: x.rolling(4).B.mean()) Out[9]: A 1 0 NaN 1 NaN 2 NaN 3 1.5 4 2.5 5 3.5 6 4.5 ... 3 33 2016-01-03 1 5 2016-01-10 1 6 2016-01-17 2 7 2016-01-24 2 8 In [13]: df.groupby('group').apply(lambda x: x.resample('1D').ffill()) Out[13]: group val group date 1 2016-01-03 1 5 2016-01-04 1 5 2016-01-05 DataFrame({'A': [1, 2, 3], ....: 'B': [4, 5, 6], ....: 'C': [7, 8, 9]}) ....: In [16]: df.where(lambda x: x > 4, lambda x: x + 10) Out[16]: 46 Chapter 1. What’s New pandas: powerful Python data analysis toolkit

data analysis toolkit, Release 0.19.1 39 3 39 [40 rows x 2 columns] In [9]: df.groupby('A').apply(lambda x: x.rolling(4).B.mean()) Out[9]: A 1 0 NaN 1 NaN 2 NaN 3 1.5 4 2.5 5 3.5 6 4.5 ... 3 33 2016-01-03 1 5 2016-01-10 1 6 2016-01-17 2 7 2016-01-24 2 8 In [13]: df.groupby('group').apply(lambda x: x.resample('1D').ffill()) Out[13]: group val group date 1 2016-01-03 1 5 2016-01-04 1 5 2016-01-05 DataFrame({'A': [1, 2, 3], ....: 'B': [4, 5, 6], ....: 'C': [7, 8, 9]}) ....: In [16]: df.where(lambda x: x > 4, lambda x: x + 10) Out[16]: A B C 0 11 14 7 1 12 5 8 2 13 6 9 .loc[], .iloc[], .ix[] These

pandas 0.25.x >>> df.resample("2D").agg(lambda x: 'a').A.dtype CategoricalDtype(categories=['a', 'b'], ordered=False) pandas 1.0.0 In [50]: df.resample("2D").agg(lambda x: 'a').A.dtype Out[50]: dtype('O') the original dtype. pandas 0.25.x >>> df.resample("2D").agg(lambda x: 'c') A 0 NaN pandas 1.0.0 In [51]: df.resample("2D").agg(lambda x: 'c') Out[51]: A 2000-01-01 c [1 rows x 1 columns] 1.5.11 by column name when axis=1 (GH27614) • Bug in core.groupby.DataFrameGroupby.agg() not able to use lambda function with named aggre- gation (GH27519) • Bug in DataFrame.groupby() losing column name information