[Tutor] Adding a new row to the dataframe with datetime as index
Hi, I am trying to add a new row to a new date in the dataframe like below df.loc['2018-01-24'] = [0,1,2,3,4,5] And I am getting the following error ValueError: cannot set using a list-like indexer with a different length than the value I do have the right number of columns and I can lookup a row by the date df.loc['2018-01-23'] df.shape (8034, 6) df.index DatetimeIndex(['2018-01-23', '2018-01-22', '2018-01-19', '2018-01-18', '2018-01-17', '2018-01-16', '2018-01-12', '2018-01-11', '2018-01-10', '2018-01-09', ... '1986-03-25', '1986-03-24', '1986-03-21', '1986-03-20', '1986-03-19', '1986-03-18', '1986-03-17', '1986-03-14', '1986-03-13', '2018-01-24'], dtype='datetime64[ns]', name='date', length=8034, freq=None) Any idea how to add a new row to a new date? -- Asif Iqbal PGP Key: 0xE62693C5 KeyServer: pgp.mit.edu A: Because it messes up the order in which people normally read text. Q: Why is top-posting such a bad thing? ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Adding a new row to the dataframe with datetime as index
Asif Iqbal wrote:
> Hi,
>
> I am trying to add a new row to a new date in the dataframe like below
>
>df.loc['2018-01-24'] = [0,1,2,3,4,5]
>
> And I am getting the following error
>
> ValueError: cannot set using a list-like indexer with a different length
> than the value
>
> I do have the right number of columns and I can lookup a row by the date
>
> df.loc['2018-01-23']
>
> df.shape
> (8034, 6)
>
> df.index
> DatetimeIndex(['2018-01-23', '2018-01-22', '2018-01-19', '2018-01-18',
>'2018-01-17', '2018-01-16', '2018-01-12', '2018-01-11',
>'2018-01-10', '2018-01-09',
>...
>'1986-03-25', '1986-03-24', '1986-03-21', '1986-03-20',
>'1986-03-19', '1986-03-18', '1986-03-17', '1986-03-14',
>'1986-03-13', '2018-01-24'],
> dtype='datetime64[ns]', name='date', length=8034, freq=None)
>
> Any idea how to add a new row to a new date?
My experiments indicate that there may be multiple values with the same key:
> >>> import pandas as pd
>>> df = pd.DataFrame([[1,2], [3,4], [5,6], [7,8]], index=["a", "b", "a",
"a"])
>>> df.loc["a"]
0 1
a 1 2
a 5 6
a 7 8
[3 rows x 2 columns]
>>> df.loc["a"] = [10, 20]
Traceback (most recent call last):
File "", line 1, in
File "/usr/lib/python3/dist-packages/pandas/core/indexing.py", line 98, in
__setitem__
self._setitem_with_indexer(indexer, value)
File "/usr/lib/python3/dist-packages/pandas/core/indexing.py", line 422,
in _setitem_with_indexer
self.obj._data = self.obj._data.setitem(indexer, value)
File "/usr/lib/python3/dist-packages/pandas/core/internals.py", line 2396,
in setitem
return self.apply('setitem', *args, **kwargs)
File "/usr/lib/python3/dist-packages/pandas/core/internals.py", line 2376,
in apply
applied = getattr(blk, f)(*args, **kwargs)
File "/usr/lib/python3/dist-packages/pandas/core/internals.py", line 615,
in setitem
raise ValueError("cannot set using a list-like indexer "
ValueError: cannot set using a list-like indexer with a different length
than the value
If found two ways to resolve this,
(1) the obvious, ensure that the lengths are the same:
>>> df.loc["a"] = [[10, 20], [30, 40], [50, 60]]
>>> df
0 1
a 10 20
b 3 4
a 30 40
a 50 60
(2) pass the key as a tuple:
>>> df.loc["a",] = [1000, 2000]
>>> df
0 1
a 1000 2000
b 3 4
a 1000 2000
a 1000 2000
[4 rows x 2 columns]
I suspect that you want neither, and instead avoid duplicate keys.
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Re: [Tutor] Adding a new row to the dataframe with datetime as index
On Mon, May 21, 2018 at 9:28 AM, Peter Otten <__pete...@web.de> wrote:
> Asif Iqbal wrote:
>
> > Hi,
> >
> > I am trying to add a new row to a new date in the dataframe like below
> >
> >df.loc['2018-01-24'] = [0,1,2,3,4,5]
> >
> > And I am getting the following error
> >
> > ValueError: cannot set using a list-like indexer with a different
> length
> > than the value
> >
> > I do have the right number of columns and I can lookup a row by the date
> >
> > df.loc['2018-01-23']
> >
> > df.shape
> > (8034, 6)
> >
> > df.index
> > DatetimeIndex(['2018-01-23', '2018-01-22', '2018-01-19', '2018-01-18',
> >'2018-01-17', '2018-01-16', '2018-01-12', '2018-01-11',
> >'2018-01-10', '2018-01-09',
> >...
> >'1986-03-25', '1986-03-24', '1986-03-21', '1986-03-20',
> >'1986-03-19', '1986-03-18', '1986-03-17', '1986-03-14',
> >'1986-03-13', '2018-01-24'],
> > dtype='datetime64[ns]', name='date', length=8034,
> freq=None)
> >
> > Any idea how to add a new row to a new date?
>
> My experiments indicate that there may be multiple values with the same
> key:
>
> > >>> import pandas as pd
> >>> df = pd.DataFrame([[1,2], [3,4], [5,6], [7,8]], index=["a", "b", "a",
> "a"])
> >>> df.loc["a"]
>0 1
> a 1 2
> a 5 6
> a 7 8
>
> [3 rows x 2 columns]
> >>> df.loc["a"] = [10, 20]
> Traceback (most recent call last):
> File "", line 1, in
> File "/usr/lib/python3/dist-packages/pandas/core/indexing.py", line 98,
> in
> __setitem__
> self._setitem_with_indexer(indexer, value)
> File "/usr/lib/python3/dist-packages/pandas/core/indexing.py", line
> 422,
> in _setitem_with_indexer
> self.obj._data = self.obj._data.setitem(indexer, value)
> File "/usr/lib/python3/dist-packages/pandas/core/internals.py", line
> 2396,
> in setitem
> return self.apply('setitem', *args, **kwargs)
> File "/usr/lib/python3/dist-packages/pandas/core/internals.py", line
> 2376,
> in apply
> applied = getattr(blk, f)(*args, **kwargs)
> File "/usr/lib/python3/dist-packages/pandas/core/internals.py", line
> 615,
> in setitem
> raise ValueError("cannot set using a list-like indexer "
> ValueError: cannot set using a list-like indexer with a different length
> than the value
>
> If found two ways to resolve this,
>
> (1) the obvious, ensure that the lengths are the same:
>
> >>> df.loc["a"] = [[10, 20], [30, 40], [50, 60]]
> >>> df
> 0 1
> a 10 20
> b 3 4
> a 30 40
> a 50 60
>
> (2) pass the key as a tuple:
>
> >>> df.loc["a",] = [1000, 2000]
> >>> df
> 0 1
> a 1000 2000
> b 3 4
> a 1000 2000
> a 1000 2000
>
> [4 rows x 2 columns]
>
> I suspect that you want neither, and instead avoid duplicate keys.
I want to overwrite the row
print ( df.loc['2018-01-24'] )
2018-01-24 0.0 1.0 2.0 3.0 4.0 NaN
df.loc['2018-01-24'] = [0,1,2,3,4,5]
ValueError: cannot set using a list-like indexer with a different length
than the value
>
> ___
> Tutor maillist - Tutor@python.org
> To unsubscribe or change subscription options:
> https://mail.python.org/mailman/listinfo/tutor
>
--
Asif Iqbal
PGP Key: 0xE62693C5 KeyServer: pgp.mit.edu
A: Because it messes up the order in which people normally read text.
Q: Why is top-posting such a bad thing?
___
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Re: [Tutor] Adding a new row to the dataframe with datetime as index
On 05/21/2018 11:17 AM, Asif Iqbal wrote: > I want to overwrite the row > > print ( df.loc['2018-01-24'] ) > 2018-01-24 0.0 1.0 2.0 3.0 4.0 NaN > > > df.loc['2018-01-24'] = [0,1,2,3,4,5] > ValueError: cannot set using a list-like indexer with a different length > than the value given what you have reported above, your current entry contains the date, so a list of seven entries; you're trying to replace it with a list of six entries, so the error seems as expected. ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Adding a new row to the dataframe with datetime as index
Asif Iqbal wrote:
> On Mon, May 21, 2018 at 9:28 AM, Peter Otten <__pete...@web.de> wrote:
>
>> Asif Iqbal wrote:
>>
>> > Hi,
>> >
>> > I am trying to add a new row to a new date in the dataframe like below
>> >
>> >df.loc['2018-01-24'] = [0,1,2,3,4,5]
>> >
>> > And I am getting the following error
>> >
>> > ValueError: cannot set using a list-like indexer with a different
>> length
>> > than the value
>> >
>> > I do have the right number of columns and I can lookup a row by the
>> > date
>> >
>> > df.loc['2018-01-23']
>> >
>> > df.shape
>> > (8034, 6)
>> >
>> > df.index
>> > DatetimeIndex(['2018-01-23', '2018-01-22', '2018-01-19',
>> > '2018-01-18',
>> >'2018-01-17', '2018-01-16', '2018-01-12', '2018-01-11',
>> >'2018-01-10', '2018-01-09',
>> >...
>> >'1986-03-25', '1986-03-24', '1986-03-21', '1986-03-20',
>> >'1986-03-19', '1986-03-18', '1986-03-17', '1986-03-14',
>> >'1986-03-13', '2018-01-24'],
>> > dtype='datetime64[ns]', name='date', length=8034,
>> freq=None)
>> >
>> > Any idea how to add a new row to a new date?
>>
>> My experiments indicate that there may be multiple values with the same
>> key:
>>
>> > >>> import pandas as pd
>> >>> df = pd.DataFrame([[1,2], [3,4], [5,6], [7,8]], index=["a", "b", "a",
>> "a"])
>> >>> df.loc["a"]
>>0 1
>> a 1 2
>> a 5 6
>> a 7 8
>>
>> [3 rows x 2 columns]
>> >>> df.loc["a"] = [10, 20]
>> Traceback (most recent call last):
>> File "", line 1, in
>> File "/usr/lib/python3/dist-packages/pandas/core/indexing.py", line 98,
>> in
>> __setitem__
>> self._setitem_with_indexer(indexer, value)
>> File "/usr/lib/python3/dist-packages/pandas/core/indexing.py", line
>> 422,
>> in _setitem_with_indexer
>> self.obj._data = self.obj._data.setitem(indexer, value)
>> File "/usr/lib/python3/dist-packages/pandas/core/internals.py", line
>> 2396,
>> in setitem
>> return self.apply('setitem', *args, **kwargs)
>> File "/usr/lib/python3/dist-packages/pandas/core/internals.py", line
>> 2376,
>> in apply
>> applied = getattr(blk, f)(*args, **kwargs)
>> File "/usr/lib/python3/dist-packages/pandas/core/internals.py", line
>> 615,
>> in setitem
>> raise ValueError("cannot set using a list-like indexer "
>> ValueError: cannot set using a list-like indexer with a different length
>> than the value
>>
>> If found two ways to resolve this,
>>
>> (1) the obvious, ensure that the lengths are the same:
>>
>> >>> df.loc["a"] = [[10, 20], [30, 40], [50, 60]]
>> >>> df
>> 0 1
>> a 10 20
>> b 3 4
>> a 30 40
>> a 50 60
>>
>> (2) pass the key as a tuple:
>>
>> >>> df.loc["a",] = [1000, 2000]
>> >>> df
>> 0 1
>> a 1000 2000
>> b 3 4
>> a 1000 2000
>> a 1000 2000
>>
>> [4 rows x 2 columns]
>>
>> I suspect that you want neither, and instead avoid duplicate keys.
>
>
>
> I want to overwrite the row
>
> print ( df.loc['2018-01-24'] )
> 2018-01-24 0.0 1.0 2.0 3.0 4.0 NaN
>
>
> df.loc['2018-01-24'] = [0,1,2,3,4,5]
> ValueError: cannot set using a list-like indexer with a different length
> than the value
Can you post a self-contained example, i. e. a small script that also
creates a -- hopefully small -- DataFrame and then triggers the ValueError?
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Re: [Tutor] Adding a new row to the dataframe with datetime as index
On Mon, May 21, 2018 at 4:59 PM, Peter Otten <__pete...@web.de> wrote:
> Asif Iqbal wrote:
>
> > On Mon, May 21, 2018 at 9:28 AM, Peter Otten <__pete...@web.de> wrote:
> >
> >> Asif Iqbal wrote:
> >>
> >> > Hi,
> >> >
> >> > I am trying to add a new row to a new date in the dataframe like below
> >> >
> >> >df.loc['2018-01-24'] = [0,1,2,3,4,5]
> >> >
> >> > And I am getting the following error
> >> >
> >> > ValueError: cannot set using a list-like indexer with a different
> >> length
> >> > than the value
> >> >
> >> > I do have the right number of columns and I can lookup a row by the
> >> > date
> >> >
> >> > df.loc['2018-01-23']
> >> >
> >> > df.shape
> >> > (8034, 6)
> >> >
> >> > df.index
> >> > DatetimeIndex(['2018-01-23', '2018-01-22', '2018-01-19',
> >> > '2018-01-18',
> >> >'2018-01-17', '2018-01-16', '2018-01-12', '2018-01-11',
> >> >'2018-01-10', '2018-01-09',
> >> >...
> >> >'1986-03-25', '1986-03-24', '1986-03-21', '1986-03-20',
> >> >'1986-03-19', '1986-03-18', '1986-03-17', '1986-03-14',
> >> >'1986-03-13', '2018-01-24'],
> >> > dtype='datetime64[ns]', name='date', length=8034,
> >> freq=None)
> >> >
> >> > Any idea how to add a new row to a new date?
> >>
> >> My experiments indicate that there may be multiple values with the same
> >> key:
> >>
> >> > >>> import pandas as pd
> >> >>> df = pd.DataFrame([[1,2], [3,4], [5,6], [7,8]], index=["a", "b",
> "a",
> >> "a"])
> >> >>> df.loc["a"]
> >>0 1
> >> a 1 2
> >> a 5 6
> >> a 7 8
> >>
> >> [3 rows x 2 columns]
> >> >>> df.loc["a"] = [10, 20]
> >> Traceback (most recent call last):
> >> File "", line 1, in
> >> File "/usr/lib/python3/dist-packages/pandas/core/indexing.py", line
> 98,
> >> in
> >> __setitem__
> >> self._setitem_with_indexer(indexer, value)
> >> File "/usr/lib/python3/dist-packages/pandas/core/indexing.py", line
> >> 422,
> >> in _setitem_with_indexer
> >> self.obj._data = self.obj._data.setitem(indexer, value)
> >> File "/usr/lib/python3/dist-packages/pandas/core/internals.py", line
> >> 2396,
> >> in setitem
> >> return self.apply('setitem', *args, **kwargs)
> >> File "/usr/lib/python3/dist-packages/pandas/core/internals.py", line
> >> 2376,
> >> in apply
> >> applied = getattr(blk, f)(*args, **kwargs)
> >> File "/usr/lib/python3/dist-packages/pandas/core/internals.py", line
> >> 615,
> >> in setitem
> >> raise ValueError("cannot set using a list-like indexer "
> >> ValueError: cannot set using a list-like indexer with a different length
> >> than the value
> >>
> >> If found two ways to resolve this,
> >>
> >> (1) the obvious, ensure that the lengths are the same:
> >>
> >> >>> df.loc["a"] = [[10, 20], [30, 40], [50, 60]]
> >> >>> df
> >> 0 1
> >> a 10 20
> >> b 3 4
> >> a 30 40
> >> a 50 60
> >>
> >> (2) pass the key as a tuple:
> >>
> >> >>> df.loc["a",] = [1000, 2000]
> >> >>> df
> >> 0 1
> >> a 1000 2000
> >> b 3 4
> >> a 1000 2000
> >> a 1000 2000
> >>
> >> [4 rows x 2 columns]
> >>
> >> I suspect that you want neither, and instead avoid duplicate keys.
> >
> >
> >
> > I want to overwrite the row
> >
> > print ( df.loc['2018-01-24'] )
> > 2018-01-24 0.0 1.0 2.0 3.0 4.0 NaN
> >
> >
> > df.loc['2018-01-24'] = [0,1,2,3,4,5]
> > ValueError: cannot set using a list-like indexer with a different
> length
> > than the value
>
> Can you post a self-contained example, i. e. a small script that also
> creates a -- hopefully small -- DataFrame and then triggers the ValueError?
>
>
It is working after I ran a df = df.sort_index()
I was looping through new dates and feeding predicted data to new row for
next day, but I was going in the wrong direction.
--
Asif Iqbal
PGP Key: 0xE62693C5 KeyServer: pgp.mit.edu
A: Because it messes up the order in which people normally read text.
Q: Why is top-posting such a bad thing?
___
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