Make Datetime Series from separate year, month, and date columns in Pandas
How can we use columns 'Yr', 'Mo' and 'Dy' to create a new column with type Datetime and set it as the index of the Pandas DataFrame?
python pandas datetime dataframe
add a comment |
How can we use columns 'Yr', 'Mo' and 'Dy' to create a new column with type Datetime and set it as the index of the Pandas DataFrame?
python pandas datetime dataframe
add a comment |
How can we use columns 'Yr', 'Mo' and 'Dy' to create a new column with type Datetime and set it as the index of the Pandas DataFrame?
python pandas datetime dataframe
How can we use columns 'Yr', 'Mo' and 'Dy' to create a new column with type Datetime and set it as the index of the Pandas DataFrame?
python pandas datetime dataframe
python pandas datetime dataframe
edited Nov 23 '18 at 13:49
Brad Solomon
13.8k83486
13.8k83486
asked Nov 20 '18 at 23:47
Shivam SinhaShivam Sinha
635
635
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
First, you should convert Yr
to a four-digit int, i.e. 1961 or 2061. This is unambiguous and, if you use the approach below, the format YYYY-MM-DD is required. That's because Pandas uses format='%Y%m%d'
in pandas/core/tools/datetimes.py:
# From pandas/core/tools/datetimes.py, if you pass a DataFrame or dict
values = to_datetime(values, format='%Y%m%d', errors=errors)
So, to take an example:
from itertools import product
import numpy as np
import pandas as pd
np.random.seed(444)
datecols = ['Yr', 'Mo', 'Dy']
mapper = dict(zip(datecols, ('year', 'month', 'day')))
df = pd.DataFrame(list(product([61, 62], [1, 2], [1, 2, 3])),
columns=datecols)
df['data'] = np.random.randn(len(df))
Here is df
:
In [11]: df
Out[11]:
Yr Mo Dy data
0 61 1 1 0.357440
1 61 1 2 0.377538
2 61 1 3 1.382338
3 61 2 1 1.175549
4 61 2 2 -0.939276
5 61 2 3 -1.143150
6 62 1 1 -0.542440
7 62 1 2 -0.548708
8 62 1 3 0.208520
9 62 2 1 0.212690
10 62 2 2 1.268021
11 62 2 3 -0.807303
Let's assume for the sake of simplicity that the true range is 1920 onward, i.e.:
In [16]: yr = df['Yr']
In [17]: df['Yr'] = np.where(yr <= 20, 2000 + yr, 1900 + yr)
In [18]: df
Out[18]:
Yr Mo Dy data
0 1961 1 1 0.357440
1 1961 1 2 0.377538
2 1961 1 3 1.382338
3 1961 2 1 1.175549
4 1961 2 2 -0.939276
5 1961 2 3 -1.143150
6 1962 1 1 -0.542440
7 1962 1 2 -0.548708
8 1962 1 3 0.208520
9 1962 2 1 0.212690
10 1962 2 2 1.268021
11 1962 2 3 -0.807303
The second thing you need to do is rename the columns; Pandas is fairly strict about this if you pass in a mapping or DataFrame to pd.to_datetime()
. Here is that step and the result:
In [21]: df.index = pd.to_datetime(df[datecols].rename(columns=mapper))
In [22]: df
Out[22]:
Yr Mo Dy data
1961-01-01 1961 1 1 0.357440
1961-01-02 1961 1 2 0.377538
1961-01-03 1961 1 3 1.382338
1961-02-01 1961 2 1 1.175549
1961-02-02 1961 2 2 -0.939276
1961-02-03 1961 2 3 -1.143150
1962-01-01 1962 1 1 -0.542440
1962-01-02 1962 1 2 -0.548708
1962-01-03 1962 1 3 0.208520
1962-02-01 1962 2 1 0.212690
1962-02-02 1962 2 2 1.268021
1962-02-03 1962 2 3 -0.807303
Lastly, here's one alternate through concatenating the columns as strings:
In [27]: as_str = df[datecols].astype(str)
In [30]: pd.to_datetime(
...: as_str['Yr'] + '-' + as_str['Mo'] +'-' + as_str['Dy'],
...: format='%y-%m-%d'
...: )
Out[30]:
0 2061-01-01
1 2061-01-02
2 2061-01-03
3 2061-02-01
4 2061-02-02
5 2061-02-03
6 2062-01-01
7 2062-01-02
8 2062-01-03
9 2062-02-01
10 2062-02-02
11 2062-02-03
dtype: datetime64[ns]
Notice again that this will assume the century for you. If you want to be explicit, you need to follow the same approach as above for adding the correct century before defining as_str
.
add a comment |
As pointed out by Brad, this is how I fixed it
def adjustyear(x):
if x >= 1800:
x = 1900 + x
else:
x = 2000 + x
return x
def parsefunc(x):
yearmodified = adjustyear(x['Yr'])
print(yearmodified)
datetimestr = str(yearmodified)+str(x['Mo'])+str(x['Dy'])
return pd.to_datetime(datetimestr, format='%Y%m%d', errors='ignore')
data['newindex'] = data.apply(parsefunc, axis=1)
data.index = data['newindex']
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
First, you should convert Yr
to a four-digit int, i.e. 1961 or 2061. This is unambiguous and, if you use the approach below, the format YYYY-MM-DD is required. That's because Pandas uses format='%Y%m%d'
in pandas/core/tools/datetimes.py:
# From pandas/core/tools/datetimes.py, if you pass a DataFrame or dict
values = to_datetime(values, format='%Y%m%d', errors=errors)
So, to take an example:
from itertools import product
import numpy as np
import pandas as pd
np.random.seed(444)
datecols = ['Yr', 'Mo', 'Dy']
mapper = dict(zip(datecols, ('year', 'month', 'day')))
df = pd.DataFrame(list(product([61, 62], [1, 2], [1, 2, 3])),
columns=datecols)
df['data'] = np.random.randn(len(df))
Here is df
:
In [11]: df
Out[11]:
Yr Mo Dy data
0 61 1 1 0.357440
1 61 1 2 0.377538
2 61 1 3 1.382338
3 61 2 1 1.175549
4 61 2 2 -0.939276
5 61 2 3 -1.143150
6 62 1 1 -0.542440
7 62 1 2 -0.548708
8 62 1 3 0.208520
9 62 2 1 0.212690
10 62 2 2 1.268021
11 62 2 3 -0.807303
Let's assume for the sake of simplicity that the true range is 1920 onward, i.e.:
In [16]: yr = df['Yr']
In [17]: df['Yr'] = np.where(yr <= 20, 2000 + yr, 1900 + yr)
In [18]: df
Out[18]:
Yr Mo Dy data
0 1961 1 1 0.357440
1 1961 1 2 0.377538
2 1961 1 3 1.382338
3 1961 2 1 1.175549
4 1961 2 2 -0.939276
5 1961 2 3 -1.143150
6 1962 1 1 -0.542440
7 1962 1 2 -0.548708
8 1962 1 3 0.208520
9 1962 2 1 0.212690
10 1962 2 2 1.268021
11 1962 2 3 -0.807303
The second thing you need to do is rename the columns; Pandas is fairly strict about this if you pass in a mapping or DataFrame to pd.to_datetime()
. Here is that step and the result:
In [21]: df.index = pd.to_datetime(df[datecols].rename(columns=mapper))
In [22]: df
Out[22]:
Yr Mo Dy data
1961-01-01 1961 1 1 0.357440
1961-01-02 1961 1 2 0.377538
1961-01-03 1961 1 3 1.382338
1961-02-01 1961 2 1 1.175549
1961-02-02 1961 2 2 -0.939276
1961-02-03 1961 2 3 -1.143150
1962-01-01 1962 1 1 -0.542440
1962-01-02 1962 1 2 -0.548708
1962-01-03 1962 1 3 0.208520
1962-02-01 1962 2 1 0.212690
1962-02-02 1962 2 2 1.268021
1962-02-03 1962 2 3 -0.807303
Lastly, here's one alternate through concatenating the columns as strings:
In [27]: as_str = df[datecols].astype(str)
In [30]: pd.to_datetime(
...: as_str['Yr'] + '-' + as_str['Mo'] +'-' + as_str['Dy'],
...: format='%y-%m-%d'
...: )
Out[30]:
0 2061-01-01
1 2061-01-02
2 2061-01-03
3 2061-02-01
4 2061-02-02
5 2061-02-03
6 2062-01-01
7 2062-01-02
8 2062-01-03
9 2062-02-01
10 2062-02-02
11 2062-02-03
dtype: datetime64[ns]
Notice again that this will assume the century for you. If you want to be explicit, you need to follow the same approach as above for adding the correct century before defining as_str
.
add a comment |
First, you should convert Yr
to a four-digit int, i.e. 1961 or 2061. This is unambiguous and, if you use the approach below, the format YYYY-MM-DD is required. That's because Pandas uses format='%Y%m%d'
in pandas/core/tools/datetimes.py:
# From pandas/core/tools/datetimes.py, if you pass a DataFrame or dict
values = to_datetime(values, format='%Y%m%d', errors=errors)
So, to take an example:
from itertools import product
import numpy as np
import pandas as pd
np.random.seed(444)
datecols = ['Yr', 'Mo', 'Dy']
mapper = dict(zip(datecols, ('year', 'month', 'day')))
df = pd.DataFrame(list(product([61, 62], [1, 2], [1, 2, 3])),
columns=datecols)
df['data'] = np.random.randn(len(df))
Here is df
:
In [11]: df
Out[11]:
Yr Mo Dy data
0 61 1 1 0.357440
1 61 1 2 0.377538
2 61 1 3 1.382338
3 61 2 1 1.175549
4 61 2 2 -0.939276
5 61 2 3 -1.143150
6 62 1 1 -0.542440
7 62 1 2 -0.548708
8 62 1 3 0.208520
9 62 2 1 0.212690
10 62 2 2 1.268021
11 62 2 3 -0.807303
Let's assume for the sake of simplicity that the true range is 1920 onward, i.e.:
In [16]: yr = df['Yr']
In [17]: df['Yr'] = np.where(yr <= 20, 2000 + yr, 1900 + yr)
In [18]: df
Out[18]:
Yr Mo Dy data
0 1961 1 1 0.357440
1 1961 1 2 0.377538
2 1961 1 3 1.382338
3 1961 2 1 1.175549
4 1961 2 2 -0.939276
5 1961 2 3 -1.143150
6 1962 1 1 -0.542440
7 1962 1 2 -0.548708
8 1962 1 3 0.208520
9 1962 2 1 0.212690
10 1962 2 2 1.268021
11 1962 2 3 -0.807303
The second thing you need to do is rename the columns; Pandas is fairly strict about this if you pass in a mapping or DataFrame to pd.to_datetime()
. Here is that step and the result:
In [21]: df.index = pd.to_datetime(df[datecols].rename(columns=mapper))
In [22]: df
Out[22]:
Yr Mo Dy data
1961-01-01 1961 1 1 0.357440
1961-01-02 1961 1 2 0.377538
1961-01-03 1961 1 3 1.382338
1961-02-01 1961 2 1 1.175549
1961-02-02 1961 2 2 -0.939276
1961-02-03 1961 2 3 -1.143150
1962-01-01 1962 1 1 -0.542440
1962-01-02 1962 1 2 -0.548708
1962-01-03 1962 1 3 0.208520
1962-02-01 1962 2 1 0.212690
1962-02-02 1962 2 2 1.268021
1962-02-03 1962 2 3 -0.807303
Lastly, here's one alternate through concatenating the columns as strings:
In [27]: as_str = df[datecols].astype(str)
In [30]: pd.to_datetime(
...: as_str['Yr'] + '-' + as_str['Mo'] +'-' + as_str['Dy'],
...: format='%y-%m-%d'
...: )
Out[30]:
0 2061-01-01
1 2061-01-02
2 2061-01-03
3 2061-02-01
4 2061-02-02
5 2061-02-03
6 2062-01-01
7 2062-01-02
8 2062-01-03
9 2062-02-01
10 2062-02-02
11 2062-02-03
dtype: datetime64[ns]
Notice again that this will assume the century for you. If you want to be explicit, you need to follow the same approach as above for adding the correct century before defining as_str
.
add a comment |
First, you should convert Yr
to a four-digit int, i.e. 1961 or 2061. This is unambiguous and, if you use the approach below, the format YYYY-MM-DD is required. That's because Pandas uses format='%Y%m%d'
in pandas/core/tools/datetimes.py:
# From pandas/core/tools/datetimes.py, if you pass a DataFrame or dict
values = to_datetime(values, format='%Y%m%d', errors=errors)
So, to take an example:
from itertools import product
import numpy as np
import pandas as pd
np.random.seed(444)
datecols = ['Yr', 'Mo', 'Dy']
mapper = dict(zip(datecols, ('year', 'month', 'day')))
df = pd.DataFrame(list(product([61, 62], [1, 2], [1, 2, 3])),
columns=datecols)
df['data'] = np.random.randn(len(df))
Here is df
:
In [11]: df
Out[11]:
Yr Mo Dy data
0 61 1 1 0.357440
1 61 1 2 0.377538
2 61 1 3 1.382338
3 61 2 1 1.175549
4 61 2 2 -0.939276
5 61 2 3 -1.143150
6 62 1 1 -0.542440
7 62 1 2 -0.548708
8 62 1 3 0.208520
9 62 2 1 0.212690
10 62 2 2 1.268021
11 62 2 3 -0.807303
Let's assume for the sake of simplicity that the true range is 1920 onward, i.e.:
In [16]: yr = df['Yr']
In [17]: df['Yr'] = np.where(yr <= 20, 2000 + yr, 1900 + yr)
In [18]: df
Out[18]:
Yr Mo Dy data
0 1961 1 1 0.357440
1 1961 1 2 0.377538
2 1961 1 3 1.382338
3 1961 2 1 1.175549
4 1961 2 2 -0.939276
5 1961 2 3 -1.143150
6 1962 1 1 -0.542440
7 1962 1 2 -0.548708
8 1962 1 3 0.208520
9 1962 2 1 0.212690
10 1962 2 2 1.268021
11 1962 2 3 -0.807303
The second thing you need to do is rename the columns; Pandas is fairly strict about this if you pass in a mapping or DataFrame to pd.to_datetime()
. Here is that step and the result:
In [21]: df.index = pd.to_datetime(df[datecols].rename(columns=mapper))
In [22]: df
Out[22]:
Yr Mo Dy data
1961-01-01 1961 1 1 0.357440
1961-01-02 1961 1 2 0.377538
1961-01-03 1961 1 3 1.382338
1961-02-01 1961 2 1 1.175549
1961-02-02 1961 2 2 -0.939276
1961-02-03 1961 2 3 -1.143150
1962-01-01 1962 1 1 -0.542440
1962-01-02 1962 1 2 -0.548708
1962-01-03 1962 1 3 0.208520
1962-02-01 1962 2 1 0.212690
1962-02-02 1962 2 2 1.268021
1962-02-03 1962 2 3 -0.807303
Lastly, here's one alternate through concatenating the columns as strings:
In [27]: as_str = df[datecols].astype(str)
In [30]: pd.to_datetime(
...: as_str['Yr'] + '-' + as_str['Mo'] +'-' + as_str['Dy'],
...: format='%y-%m-%d'
...: )
Out[30]:
0 2061-01-01
1 2061-01-02
2 2061-01-03
3 2061-02-01
4 2061-02-02
5 2061-02-03
6 2062-01-01
7 2062-01-02
8 2062-01-03
9 2062-02-01
10 2062-02-02
11 2062-02-03
dtype: datetime64[ns]
Notice again that this will assume the century for you. If you want to be explicit, you need to follow the same approach as above for adding the correct century before defining as_str
.
First, you should convert Yr
to a four-digit int, i.e. 1961 or 2061. This is unambiguous and, if you use the approach below, the format YYYY-MM-DD is required. That's because Pandas uses format='%Y%m%d'
in pandas/core/tools/datetimes.py:
# From pandas/core/tools/datetimes.py, if you pass a DataFrame or dict
values = to_datetime(values, format='%Y%m%d', errors=errors)
So, to take an example:
from itertools import product
import numpy as np
import pandas as pd
np.random.seed(444)
datecols = ['Yr', 'Mo', 'Dy']
mapper = dict(zip(datecols, ('year', 'month', 'day')))
df = pd.DataFrame(list(product([61, 62], [1, 2], [1, 2, 3])),
columns=datecols)
df['data'] = np.random.randn(len(df))
Here is df
:
In [11]: df
Out[11]:
Yr Mo Dy data
0 61 1 1 0.357440
1 61 1 2 0.377538
2 61 1 3 1.382338
3 61 2 1 1.175549
4 61 2 2 -0.939276
5 61 2 3 -1.143150
6 62 1 1 -0.542440
7 62 1 2 -0.548708
8 62 1 3 0.208520
9 62 2 1 0.212690
10 62 2 2 1.268021
11 62 2 3 -0.807303
Let's assume for the sake of simplicity that the true range is 1920 onward, i.e.:
In [16]: yr = df['Yr']
In [17]: df['Yr'] = np.where(yr <= 20, 2000 + yr, 1900 + yr)
In [18]: df
Out[18]:
Yr Mo Dy data
0 1961 1 1 0.357440
1 1961 1 2 0.377538
2 1961 1 3 1.382338
3 1961 2 1 1.175549
4 1961 2 2 -0.939276
5 1961 2 3 -1.143150
6 1962 1 1 -0.542440
7 1962 1 2 -0.548708
8 1962 1 3 0.208520
9 1962 2 1 0.212690
10 1962 2 2 1.268021
11 1962 2 3 -0.807303
The second thing you need to do is rename the columns; Pandas is fairly strict about this if you pass in a mapping or DataFrame to pd.to_datetime()
. Here is that step and the result:
In [21]: df.index = pd.to_datetime(df[datecols].rename(columns=mapper))
In [22]: df
Out[22]:
Yr Mo Dy data
1961-01-01 1961 1 1 0.357440
1961-01-02 1961 1 2 0.377538
1961-01-03 1961 1 3 1.382338
1961-02-01 1961 2 1 1.175549
1961-02-02 1961 2 2 -0.939276
1961-02-03 1961 2 3 -1.143150
1962-01-01 1962 1 1 -0.542440
1962-01-02 1962 1 2 -0.548708
1962-01-03 1962 1 3 0.208520
1962-02-01 1962 2 1 0.212690
1962-02-02 1962 2 2 1.268021
1962-02-03 1962 2 3 -0.807303
Lastly, here's one alternate through concatenating the columns as strings:
In [27]: as_str = df[datecols].astype(str)
In [30]: pd.to_datetime(
...: as_str['Yr'] + '-' + as_str['Mo'] +'-' + as_str['Dy'],
...: format='%y-%m-%d'
...: )
Out[30]:
0 2061-01-01
1 2061-01-02
2 2061-01-03
3 2061-02-01
4 2061-02-02
5 2061-02-03
6 2062-01-01
7 2062-01-02
8 2062-01-03
9 2062-02-01
10 2062-02-02
11 2062-02-03
dtype: datetime64[ns]
Notice again that this will assume the century for you. If you want to be explicit, you need to follow the same approach as above for adding the correct century before defining as_str
.
edited Nov 21 '18 at 0:15
answered Nov 21 '18 at 0:05
Brad SolomonBrad Solomon
13.8k83486
13.8k83486
add a comment |
add a comment |
As pointed out by Brad, this is how I fixed it
def adjustyear(x):
if x >= 1800:
x = 1900 + x
else:
x = 2000 + x
return x
def parsefunc(x):
yearmodified = adjustyear(x['Yr'])
print(yearmodified)
datetimestr = str(yearmodified)+str(x['Mo'])+str(x['Dy'])
return pd.to_datetime(datetimestr, format='%Y%m%d', errors='ignore')
data['newindex'] = data.apply(parsefunc, axis=1)
data.index = data['newindex']
add a comment |
As pointed out by Brad, this is how I fixed it
def adjustyear(x):
if x >= 1800:
x = 1900 + x
else:
x = 2000 + x
return x
def parsefunc(x):
yearmodified = adjustyear(x['Yr'])
print(yearmodified)
datetimestr = str(yearmodified)+str(x['Mo'])+str(x['Dy'])
return pd.to_datetime(datetimestr, format='%Y%m%d', errors='ignore')
data['newindex'] = data.apply(parsefunc, axis=1)
data.index = data['newindex']
add a comment |
As pointed out by Brad, this is how I fixed it
def adjustyear(x):
if x >= 1800:
x = 1900 + x
else:
x = 2000 + x
return x
def parsefunc(x):
yearmodified = adjustyear(x['Yr'])
print(yearmodified)
datetimestr = str(yearmodified)+str(x['Mo'])+str(x['Dy'])
return pd.to_datetime(datetimestr, format='%Y%m%d', errors='ignore')
data['newindex'] = data.apply(parsefunc, axis=1)
data.index = data['newindex']
As pointed out by Brad, this is how I fixed it
def adjustyear(x):
if x >= 1800:
x = 1900 + x
else:
x = 2000 + x
return x
def parsefunc(x):
yearmodified = adjustyear(x['Yr'])
print(yearmodified)
datetimestr = str(yearmodified)+str(x['Mo'])+str(x['Dy'])
return pd.to_datetime(datetimestr, format='%Y%m%d', errors='ignore')
data['newindex'] = data.apply(parsefunc, axis=1)
data.index = data['newindex']
answered Nov 21 '18 at 5:36
Shivam SinhaShivam Sinha
635
635
add a comment |
add a comment |
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StackExchange.helpers.onClickDraftSave('#login-link');
});
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown