BUG: pd.period_range ignores multiple of start frequency

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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

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BUG: `pd.period_range` ignores multiple of `start` frequency #47465

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	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

Uh oh!

BUG: pd.period_range ignores multiple of start frequency #47465

Description

Pandas version checks

Reproducible Example

Issue Description

Expected Behavior

Installed Versions

Metadata

Metadata

Assignees

Labels

Type

Projects

Milestone

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Development

Issue actions

BUG: `pd.period_range` ignores multiple of `start` frequency #47465