BUG: `pd.period_range` ignores multiple of `start` frequency

[jaheba]

Pandas version checks

• I have checked that this issue has not already been reported.

• I have confirmed this bug exists on the latest version of pandas.

• I have confirmed this bug exists on the main branch of pandas.

Reproducible Example

import pandas as pd

start = pd.Period("2020", freq="3D")

a = pd.period_range(start, periods=2, freq=start.freq)
assert a[1] == start + 1

# Fails
b = pd.period_range(start, periods=2)
assert b[1] == start + 1

Issue Description

The issue lies in _get_ordinal_range:

pandas/pandas/core/arrays/period.py

Lines 1044 to 1089 in 37d4c31

	def _get_ordinal_range(start, end, periods, freq, mult=1):
	if com.count_not_none(start, end, periods) != 2:
	raise ValueError(
	"Of the three parameters: start, end, and periods, "
	"exactly two must be specified"
	)
	if freq is not None:
	freq = to_offset(freq)
	mult = freq.n
	if start is not None:
	start = Period(start, freq)
	if end is not None:
	end = Period(end, freq)
	is_start_per = isinstance(start, Period)
	is_end_per = isinstance(end, Period)
	if is_start_per and is_end_per and start.freq != end.freq:
	raise ValueError("start and end must have same freq")
	if start is NaT or end is NaT:
	raise ValueError("start and end must not be NaT")
	if freq is None:
	if is_start_per:
	freq = start.freq
	elif is_end_per:
	freq = end.freq
	else: # pragma: no cover
	raise ValueError("Could not infer freq from start/end")
	if periods is not None:
	periods = periods * mult
	if start is None:
	data = np.arange(
	end.ordinal - periods + mult, end.ordinal + 1, mult, dtype=np.int64
	)
	else:
	data = np.arange(
	start.ordinal, start.ordinal + periods, mult, dtype=np.int64
	)
	else:
	data = np.arange(start.ordinal, end.ordinal + 1, mult, dtype=np.int64)
	return data, freq

A fix could be to replace:

pandas/pandas/core/arrays/period.py

Lines 1068 to 1072 in 37d4c31

	if freq is None:
	if is_start_per:
	freq = start.freq
	elif is_end_per:
	freq = end.freq

with

   if freq is None:
        if is_start_per:
            freq = start.freq
            mult = freq.n
        elif is_end_per:
            freq = end.freq
            mult = freq.n

Expected Behavior

Omitting the freq argument of pd.period_range should be the same as passing the freq of the start or end field to it via freq=.

Installed Versions

Not relevant to issue.

[simonjayhawkins]

Thanks @jaheba for the report.

Expected Behavior

Omitting the freq argument of pd.period_range should be the same as passing the freq of the start or end field to it via freq=.

from the docs...

freq: str or DateOffset, optional
Frequency alias. By default the freq is taken from start or end if those are Period objects. Otherwise, the default is "D" for daily frequency.

So this does indeed appear to be a bug.

start = pd.Period("2020", freq="3D")
print(pd.period_range(start, periods=2))
# PeriodIndex(['2020-01-01', '2020-01-02'], dtype='period[3D]')
print(b.freq)
# <3 * Days>

Note : the PeriodIndex frequency is not the difference between any two consecutive elements. see #47227

but should be consistent with specifying the freq explicitly as in the OP.

start = pd.Period("2020", freq="3D")
print(start.freq)
<3 * Days>
print(pd.period_range(start, periods=2, freq=start.freq))
# PeriodIndex(['2020-01-01', '2020-01-04'], dtype='period[3D]')

contributions and PRs welcome.

[jaheba]

Thanks @simonjayhawkins -- I have submitted a PR: #47598

[natmokval]

I think, PR #53709 closes this issue.
If there is no objection, I close this issue.