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pd.Timedelta() cannot process offset alias T while pd.infer_freq() return this format. #21762

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elboerto opened this issue Jul 6, 2018 · 1 comment

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commented Jul 6, 2018

Code Sample, a copy-pastable example if possible

import pandas as pd
series = pd.Series([1, 2, 3], pd.date_range('201701010000', '201701010030', freq='15min'))
print(pd.infer_freq(series.index))
# Error
pd.Timedelta(pd.infer_freq(series.index))
# While this works
pd.Timedelta(series.index.freq)

Problem description

pd.Timedelta constructor works with e.g. '15min' but not with '15T',
while pd.infer_freq() returns '15T' and pd.Series.index.freq returns '15min'.
This leads to errors.

Expected Output

Same behaviour of pd.Timedelta() for 'min' and 'T' offset alias or deprecation of 'T' as an offset alias.

Output of pd.show_versions()

INSTALLED VERSIONS ------------------ commit: None python: 3.6.4.final.0 python-bits: 64 OS: Windows OS-release: 10 machine: AMD64 processor: Intel64 Family 6 Model 142 Stepping 9, GenuineIntel byteorder: little LC_ALL: None LANG: None LOCALE: None.None pandas: 0.23.1 pytest: 3.6.2 pip: 10.0.1 setuptools: 39.2.0 Cython: None numpy: 1.14.5 scipy: 1.1.0 pyarrow: None xarray: None IPython: 6.4.0 sphinx: None patsy: None dateutil: 2.7.3 pytz: 2018.5 blosc: None bottleneck: None tables: None numexpr: None feather: None matplotlib: 2.2.2 openpyxl: None xlrd: None xlwt: None xlsxwriter: None lxml: None bs4: None html5lib: None sqlalchemy: None pymysql: None psycopg2: None jinja2: None s3fs: None fastparquet: None pandas_gbq: None pandas_datareader: None
@gfyoung

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commented Jul 6, 2018

Weird! My first inclination is to accept T as an offset.

cc @jreback

@sinhrks sinhrks referenced this issue Nov 1, 2018

Merged

BUG: Cleanup timedelta offset #23439

4 of 4 tasks complete

@jreback jreback added this to the 0.24.0 milestone Nov 6, 2018

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