Selecting Rows with DateTimeIndex without referring to date
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Is there a way to select rows with a DateTimeIndex without referring to the date as such e.g. selecting row index 2 (the usual Python default manner) rather than "1995-02-02"?
Thanks in advance.
python python-3.x pandas
asked Nov 23 '18 at 1:52
AnalysisNerdAnalysisNerd
69111
add a comment |
Is there a way to select rows with a DateTimeIndex without referring to the date as such e.g. selecting row index 2 (the usual Python default manner) rather than "1995-02-02"?
Thanks in advance.
python python-3.x pandas
asked Nov 23 '18 at 1:52
AnalysisNerdAnalysisNerd
69111
add a comment |
Is there a way to select rows with a DateTimeIndex without referring to the date as such e.g. selecting row index 2 (the usual Python default manner) rather than "1995-02-02"?
Thanks in advance.
python python-3.x pandas
asked Nov 23 '18 at 1:52
AnalysisNerdAnalysisNerd
69111
Is there a way to select rows with a DateTimeIndex without referring to the date as such e.g. selecting row index 2 (the usual Python default manner) rather than "1995-02-02"?
Thanks in advance.
python python-3.x pandas
python python-3.x pandas
asked Nov 23 '18 at 1:52
AnalysisNerdAnalysisNerd
69111
asked Nov 23 '18 at 1:52
AnalysisNerdAnalysisNerd
69111
asked Nov 23 '18 at 1:52
AnalysisNerdAnalysisNerd
69111
asked Nov 23 '18 at 1:52
AnalysisNerdAnalysisNerd
69111
asked Nov 23 '18 at 1:52
AnalysisNerdAnalysisNerd
69111
69111
add a comment |
add a comment |
1 Answer
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Yes, you can use .iloc, the positional indexer:
df.iloc[2]
Basically, it indexes by actual position starting from 0 to len(df), allowing slicing too:
df.iloc[2:5]
It also works for columns (by position, again):
df.iloc[:, 0] # All rows, first column
df.iloc[0:2, 0:2] # First 2 rows, first 2 columns
answered Nov 23 '18 at 2:09
Julian PellerJulian Peller
9141512
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Yes, you can use .iloc, the positional indexer:
df.iloc[2]
Basically, it indexes by actual position starting from 0 to len(df), allowing slicing too:
df.iloc[2:5]
It also works for columns (by position, again):
df.iloc[:, 0] # All rows, first column
df.iloc[0:2, 0:2] # First 2 rows, first 2 columns
answered Nov 23 '18 at 2:09
Julian PellerJulian Peller
9141512
add a comment |
Yes, you can use .iloc, the positional indexer:
df.iloc[2]
Basically, it indexes by actual position starting from 0 to len(df), allowing slicing too:
df.iloc[2:5]
It also works for columns (by position, again):
df.iloc[:, 0] # All rows, first column
df.iloc[0:2, 0:2] # First 2 rows, first 2 columns
answered Nov 23 '18 at 2:09
Julian PellerJulian Peller
9141512
add a comment |
Yes, you can use .iloc, the positional indexer:
df.iloc[2]
Basically, it indexes by actual position starting from 0 to len(df), allowing slicing too:
df.iloc[2:5]
It also works for columns (by position, again):
df.iloc[:, 0] # All rows, first column
df.iloc[0:2, 0:2] # First 2 rows, first 2 columns
answered Nov 23 '18 at 2:09
Julian PellerJulian Peller
9141512
Yes, you can use .iloc, the positional indexer:
df.iloc[2]
Basically, it indexes by actual position starting from 0 to len(df), allowing slicing too:
df.iloc[2:5]
It also works for columns (by position, again):
df.iloc[:, 0] # All rows, first column
df.iloc[0:2, 0:2] # First 2 rows, first 2 columns
answered Nov 23 '18 at 2:09
Julian PellerJulian Peller
9141512
answered Nov 23 '18 at 2:09
Julian PellerJulian Peller
9141512
answered Nov 23 '18 at 2:09
Julian PellerJulian Peller
9141512
answered Nov 23 '18 at 2:09
Julian PellerJulian Peller
9141512
9141512
add a comment |
add a comment |
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