How to convert a series of tuples into a pandas dataframe?
1
Assume that we have the following pandas series resulted from an apply function applied on a dataframe after groupby.
<class 'pandas.core.series.Series'>
0 (1, 0, [0.2, 0.2, 0.2], [0.2, 0.2, 0.2])
1 (2, 1000, [0.6, 0.7, 0.5], [0.1, 0.3, 0.1])
2 (1, 0, [0.4, 0.4, 0.4], [0.4, 0.4, 0.4])
3 (1, 0, [0.5, 0.5, 0.5], [0.5, 0.5, 0.5])
4 (3, 14000, [0.8, 0.8, 0.8], [0.6, 0.6, 0.6])
dtype: object
Can we convert this into a dataframe when the sigList=['sig1','sig2', 'sig3'] are given?
Length Distance sig1Max sig2Max sig3Max sig1Min sig2Min sig3Min
1 0 0.2 0.2 0.2 0.2 0.2 0.2
2 1000 0.6 0.7 0.5 0.1 0.3 0.1
1 0 0.4 0.4 0.4 0.4 0.4 0.4
1 0 0.5 0.5 0.5 0.5 0.5 0.5
3 14000 0.8 0.8 0.8 0.6 0.6 0.6
Thanks in advance
python pandas dataframe series
asked Nov 20 '18 at 22:32
burcakburcak
638
add a comment |
1
Assume that we have the following pandas series resulted from an apply function applied on a dataframe after groupby.
<class 'pandas.core.series.Series'>
0 (1, 0, [0.2, 0.2, 0.2], [0.2, 0.2, 0.2])
1 (2, 1000, [0.6, 0.7, 0.5], [0.1, 0.3, 0.1])
2 (1, 0, [0.4, 0.4, 0.4], [0.4, 0.4, 0.4])
3 (1, 0, [0.5, 0.5, 0.5], [0.5, 0.5, 0.5])
4 (3, 14000, [0.8, 0.8, 0.8], [0.6, 0.6, 0.6])
dtype: object
Can we convert this into a dataframe when the sigList=['sig1','sig2', 'sig3'] are given?
Length Distance sig1Max sig2Max sig3Max sig1Min sig2Min sig3Min
1 0 0.2 0.2 0.2 0.2 0.2 0.2
2 1000 0.6 0.7 0.5 0.1 0.3 0.1
1 0 0.4 0.4 0.4 0.4 0.4 0.4
1 0 0.5 0.5 0.5 0.5 0.5 0.5
3 14000 0.8 0.8 0.8 0.6 0.6 0.6
Thanks in advance
python pandas dataframe series
asked Nov 20 '18 at 22:32
burcakburcak
638
def myfunc(x,signatures): return x.shape[0], x['start'].iloc[-1] - x['start'].iloc[0], x[signatures].agg(max), x[signatures].agg(min)This function creates the lists in tuples. Can we flatten the list in the return statement.
– burcak
Nov 20 '18 at 22:59
add a comment |
1
1
1
Assume that we have the following pandas series resulted from an apply function applied on a dataframe after groupby.
<class 'pandas.core.series.Series'>
0 (1, 0, [0.2, 0.2, 0.2], [0.2, 0.2, 0.2])
1 (2, 1000, [0.6, 0.7, 0.5], [0.1, 0.3, 0.1])
2 (1, 0, [0.4, 0.4, 0.4], [0.4, 0.4, 0.4])
3 (1, 0, [0.5, 0.5, 0.5], [0.5, 0.5, 0.5])
4 (3, 14000, [0.8, 0.8, 0.8], [0.6, 0.6, 0.6])
dtype: object
Can we convert this into a dataframe when the sigList=['sig1','sig2', 'sig3'] are given?
Length Distance sig1Max sig2Max sig3Max sig1Min sig2Min sig3Min
1 0 0.2 0.2 0.2 0.2 0.2 0.2
2 1000 0.6 0.7 0.5 0.1 0.3 0.1
1 0 0.4 0.4 0.4 0.4 0.4 0.4
1 0 0.5 0.5 0.5 0.5 0.5 0.5
3 14000 0.8 0.8 0.8 0.6 0.6 0.6
Thanks in advance
python pandas dataframe series
asked Nov 20 '18 at 22:32
burcakburcak
638
Assume that we have the following pandas series resulted from an apply function applied on a dataframe after groupby.
<class 'pandas.core.series.Series'>
0 (1, 0, [0.2, 0.2, 0.2], [0.2, 0.2, 0.2])
1 (2, 1000, [0.6, 0.7, 0.5], [0.1, 0.3, 0.1])
2 (1, 0, [0.4, 0.4, 0.4], [0.4, 0.4, 0.4])
3 (1, 0, [0.5, 0.5, 0.5], [0.5, 0.5, 0.5])
4 (3, 14000, [0.8, 0.8, 0.8], [0.6, 0.6, 0.6])
dtype: object
Can we convert this into a dataframe when the sigList=['sig1','sig2', 'sig3'] are given?
Length Distance sig1Max sig2Max sig3Max sig1Min sig2Min sig3Min
1 0 0.2 0.2 0.2 0.2 0.2 0.2
2 1000 0.6 0.7 0.5 0.1 0.3 0.1
1 0 0.4 0.4 0.4 0.4 0.4 0.4
1 0 0.5 0.5 0.5 0.5 0.5 0.5
3 14000 0.8 0.8 0.8 0.6 0.6 0.6
Thanks in advance
python pandas dataframe series
python pandas dataframe series
asked Nov 20 '18 at 22:32
burcakburcak
638
asked Nov 20 '18 at 22:32
burcakburcak
638
asked Nov 20 '18 at 22:32
burcakburcak
638
asked Nov 20 '18 at 22:32
burcakburcak
638
asked Nov 20 '18 at 22:32
burcakburcak
638
638
def myfunc(x,signatures): return x.shape[0], x['start'].iloc[-1] - x['start'].iloc[0], x[signatures].agg(max), x[signatures].agg(min)This function creates the lists in tuples. Can we flatten the list in the return statement.
– burcak
Nov 20 '18 at 22:59
add a comment |
def myfunc(x,signatures): return x.shape[0], x['start'].iloc[-1] - x['start'].iloc[0], x[signatures].agg(max), x[signatures].agg(min)This function creates the lists in tuples. Can we flatten the list in the return statement.
– burcak
Nov 20 '18 at 22:59
def myfunc(x,signatures): return x.shape[0], x['start'].iloc[-1] - x['start'].iloc[0], x[signatures].agg(max), x[signatures].agg(min) This function creates the lists in tuples. Can we flatten the list in the return statement.– burcak
Nov 20 '18 at 22:59
def myfunc(x,signatures): return x.shape[0], x['start'].iloc[-1] - x['start'].iloc[0], x[signatures].agg(max), x[signatures].agg(min) This function creates the lists in tuples. Can we flatten the list in the return statement.– burcak
Nov 20 '18 at 22:59
add a comment |
3 Answers
3
active
oldest
votes
3
Do it the old fashioned (and fast) way, using a list comprehension:
columns = ("Length Distance sig1Max sig2Max"
"sig3Max sig1Min sig2Min sig3Min").split()
df = pd.DataFrame([[a, b, *c, *d] for a,b,c,d in series.values], columns=columns)
print(df)
Length Distance sig1Max sig2Max sig3Max sig1Min sig2Min sig3Min
0 1 0 0.2 0.2 0.2 0.2 0.2 0.2
1 2 1000 0.6 0.7 0.5 0.1 0.3 0.1
2 1 0 0.4 0.4 0.4 0.4 0.4 0.4
3 1 0 0.5 0.5 0.5 0.5 0.5 0.5
4 3 14000 0.8 0.8 0.8 0.6 0.6 0.6
Or, perhaps you meant, do it a little more dynamically
sigList = ['sig1', 'sig2', 'sig3']
columns = ['Length', 'Distance']
columns.extend(f'{s}{lbl}' for lbl in ('Max', 'Min') for s in sigList )
df = pd.DataFrame([[a,b,*c,*d] for a,b,c,d in series.values], columns=columns)
print(df)
Length Distance sig1Max sig2Max sig3Max sig1Min sig2Min sig3Min
0 1 0 0.2 0.2 0.2 0.2 0.2 0.2
1 2 1000 0.6 0.7 0.5 0.1 0.3 0.1
2 1 0 0.4 0.4 0.4 0.4 0.4 0.4
3 1 0 0.5 0.5 0.5 0.5 0.5 0.5
4 3 14000 0.8 0.8 0.8 0.6 0.6 0.6
answered Nov 20 '18 at 23:26
coldspeedcoldspeed
135k23145231
Great, especially the dynamic one, since in my case sigList is resolved at runtime.
– burcak
Nov 21 '18 at 3:48
@burcak I had the order of the loops interchanged with the second one (sorry!) but you got the idea.
– coldspeed
Nov 21 '18 at 3:53
Thanks. In fact, the function below generates the series of tuples when called in apply method of grouped dataframe. Is there a way to just return dataframe so I won't need to create dataframe from series later on? def myfunc(x,signatures): return x.shape[0], x['start'].iloc[-1] - x['start'].iloc[0], x[signatures].agg(max), x[signatures].agg(min)
– burcak
Nov 21 '18 at 4:08
add a comment |
1
You may check
newdf=pd.DataFrame(s.tolist())
newdf=pd.concat([newdf[[0,1]],pd.DataFrame(newdf[2].tolist()),pd.DataFrame(newdf[3].tolist())],1)
newdf.columns = [
"Length", "Distance", "sig1Max", "sig2Max", "sig3Max", "sig1Min", "sig2Min", "sig3Min"
]
newdf
Out[163]:
Length Distance sig1Max ... sig1Min sig2Min sig3Min
0 1 0 0.2 ... 0.2 0.2 0.2
1 2 1000 0.6 ... 0.1 0.3 0.1
2 1 0 0.4 ... 0.4 0.4 0.4
3 1 0 0.5 ... 0.5 0.5 0.5
4 3 14000 0.8 ... 0.6 0.6 0.6
[5 rows x 8 columns]
answered Nov 20 '18 at 23:17
Wen-BenWen-Ben
116k83369
Found a way as well!
– coldspeed
Nov 20 '18 at 23:26
add a comment |
0
You can flatten each element and then convert each to a Series itself. Converting each element to a Series turns the main Series (s in the example below) into a DataFrame. Then just set the column names as you wish.
For example:
import pandas as pd
# load in your data
s = pd.Series([
(1, 0, [0.2, 0.2, 0.2], [0.2, 0.2, 0.2]),
(2, 1000, [0.6, 0.7, 0.5], [0.1, 0.3, 0.1]),
(1, 0, [0.4, 0.4, 0.4], [0.4, 0.4, 0.4]),
(1, 0, [0.5, 0.5, 0.5], [0.5, 0.5, 0.5]),
(3, 14000, [0.8, 0.8, 0.8], [0.6, 0.6, 0.6]),
])
def flatten(x):
# note this is not very robust, but works for this case
return [x[0], x[1], *x[2], *x[3]]
df = s.apply(flatten).apply(pd.Series)
df.columns = [
"Length", "Distance", "sig1Max", "sig2Max", "sig3Max", "sig1Min", "sig2Min", "sig3Min"
]
Then you have df as:
Length Distance sig1Max sig2Max sig3Max sig1Min sig2Min sig3Min
0 1.0 0.0 0.2 0.2 0.2 0.2 0.2 0.2
1 2.0 1000.0 0.6 0.7 0.5 0.1 0.3 0.1
2 1.0 0.0 0.4 0.4 0.4 0.4 0.4 0.4
3 1.0 0.0 0.5 0.5 0.5 0.5 0.5 0.5
4 3.0 14000.0 0.8 0.8 0.8 0.6 0.6 0.6
answered Nov 20 '18 at 22:41
Henry WoodyHenry Woody
4,7273927
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
Do it the old fashioned (and fast) way, using a list comprehension:
columns = ("Length Distance sig1Max sig2Max"
"sig3Max sig1Min sig2Min sig3Min").split()
df = pd.DataFrame([[a, b, *c, *d] for a,b,c,d in series.values], columns=columns)
print(df)
Length Distance sig1Max sig2Max sig3Max sig1Min sig2Min sig3Min
0 1 0 0.2 0.2 0.2 0.2 0.2 0.2
1 2 1000 0.6 0.7 0.5 0.1 0.3 0.1
2 1 0 0.4 0.4 0.4 0.4 0.4 0.4
3 1 0 0.5 0.5 0.5 0.5 0.5 0.5
4 3 14000 0.8 0.8 0.8 0.6 0.6 0.6
Or, perhaps you meant, do it a little more dynamically
sigList = ['sig1', 'sig2', 'sig3']
columns = ['Length', 'Distance']
columns.extend(f'{s}{lbl}' for lbl in ('Max', 'Min') for s in sigList )
df = pd.DataFrame([[a,b,*c,*d] for a,b,c,d in series.values], columns=columns)
print(df)
Length Distance sig1Max sig2Max sig3Max sig1Min sig2Min sig3Min
0 1 0 0.2 0.2 0.2 0.2 0.2 0.2
1 2 1000 0.6 0.7 0.5 0.1 0.3 0.1
2 1 0 0.4 0.4 0.4 0.4 0.4 0.4
3 1 0 0.5 0.5 0.5 0.5 0.5 0.5
4 3 14000 0.8 0.8 0.8 0.6 0.6 0.6
answered Nov 20 '18 at 23:26
coldspeedcoldspeed
135k23145231
Great, especially the dynamic one, since in my case sigList is resolved at runtime.
– burcak
Nov 21 '18 at 3:48
@burcak I had the order of the loops interchanged with the second one (sorry!) but you got the idea.
– coldspeed
Nov 21 '18 at 3:53
Thanks. In fact, the function below generates the series of tuples when called in apply method of grouped dataframe. Is there a way to just return dataframe so I won't need to create dataframe from series later on? def myfunc(x,signatures): return x.shape[0], x['start'].iloc[-1] - x['start'].iloc[0], x[signatures].agg(max), x[signatures].agg(min)
– burcak
Nov 21 '18 at 4:08
add a comment |
3
Do it the old fashioned (and fast) way, using a list comprehension:
columns = ("Length Distance sig1Max sig2Max"
"sig3Max sig1Min sig2Min sig3Min").split()
df = pd.DataFrame([[a, b, *c, *d] for a,b,c,d in series.values], columns=columns)
print(df)
Length Distance sig1Max sig2Max sig3Max sig1Min sig2Min sig3Min
0 1 0 0.2 0.2 0.2 0.2 0.2 0.2
1 2 1000 0.6 0.7 0.5 0.1 0.3 0.1
2 1 0 0.4 0.4 0.4 0.4 0.4 0.4
3 1 0 0.5 0.5 0.5 0.5 0.5 0.5
4 3 14000 0.8 0.8 0.8 0.6 0.6 0.6
Or, perhaps you meant, do it a little more dynamically
sigList = ['sig1', 'sig2', 'sig3']
columns = ['Length', 'Distance']
columns.extend(f'{s}{lbl}' for lbl in ('Max', 'Min') for s in sigList )
df = pd.DataFrame([[a,b,*c,*d] for a,b,c,d in series.values], columns=columns)
print(df)
Length Distance sig1Max sig2Max sig3Max sig1Min sig2Min sig3Min
0 1 0 0.2 0.2 0.2 0.2 0.2 0.2
1 2 1000 0.6 0.7 0.5 0.1 0.3 0.1
2 1 0 0.4 0.4 0.4 0.4 0.4 0.4
3 1 0 0.5 0.5 0.5 0.5 0.5 0.5
4 3 14000 0.8 0.8 0.8 0.6 0.6 0.6
answered Nov 20 '18 at 23:26
coldspeedcoldspeed
135k23145231
Great, especially the dynamic one, since in my case sigList is resolved at runtime.
– burcak
Nov 21 '18 at 3:48
@burcak I had the order of the loops interchanged with the second one (sorry!) but you got the idea.
– coldspeed
Nov 21 '18 at 3:53
Thanks. In fact, the function below generates the series of tuples when called in apply method of grouped dataframe. Is there a way to just return dataframe so I won't need to create dataframe from series later on? def myfunc(x,signatures): return x.shape[0], x['start'].iloc[-1] - x['start'].iloc[0], x[signatures].agg(max), x[signatures].agg(min)
– burcak
Nov 21 '18 at 4:08
add a comment |
3
3
3
Do it the old fashioned (and fast) way, using a list comprehension:
columns = ("Length Distance sig1Max sig2Max"
"sig3Max sig1Min sig2Min sig3Min").split()
df = pd.DataFrame([[a, b, *c, *d] for a,b,c,d in series.values], columns=columns)
print(df)
Length Distance sig1Max sig2Max sig3Max sig1Min sig2Min sig3Min
0 1 0 0.2 0.2 0.2 0.2 0.2 0.2
1 2 1000 0.6 0.7 0.5 0.1 0.3 0.1
2 1 0 0.4 0.4 0.4 0.4 0.4 0.4
3 1 0 0.5 0.5 0.5 0.5 0.5 0.5
4 3 14000 0.8 0.8 0.8 0.6 0.6 0.6
Or, perhaps you meant, do it a little more dynamically
sigList = ['sig1', 'sig2', 'sig3']
columns = ['Length', 'Distance']
columns.extend(f'{s}{lbl}' for lbl in ('Max', 'Min') for s in sigList )
df = pd.DataFrame([[a,b,*c,*d] for a,b,c,d in series.values], columns=columns)
print(df)
Length Distance sig1Max sig2Max sig3Max sig1Min sig2Min sig3Min
0 1 0 0.2 0.2 0.2 0.2 0.2 0.2
1 2 1000 0.6 0.7 0.5 0.1 0.3 0.1
2 1 0 0.4 0.4 0.4 0.4 0.4 0.4
3 1 0 0.5 0.5 0.5 0.5 0.5 0.5
4 3 14000 0.8 0.8 0.8 0.6 0.6 0.6
answered Nov 20 '18 at 23:26
coldspeedcoldspeed
135k23145231
Do it the old fashioned (and fast) way, using a list comprehension:
columns = ("Length Distance sig1Max sig2Max"
"sig3Max sig1Min sig2Min sig3Min").split()
df = pd.DataFrame([[a, b, *c, *d] for a,b,c,d in series.values], columns=columns)
print(df)
Length Distance sig1Max sig2Max sig3Max sig1Min sig2Min sig3Min
0 1 0 0.2 0.2 0.2 0.2 0.2 0.2
1 2 1000 0.6 0.7 0.5 0.1 0.3 0.1
2 1 0 0.4 0.4 0.4 0.4 0.4 0.4
3 1 0 0.5 0.5 0.5 0.5 0.5 0.5
4 3 14000 0.8 0.8 0.8 0.6 0.6 0.6
Or, perhaps you meant, do it a little more dynamically
sigList = ['sig1', 'sig2', 'sig3']
columns = ['Length', 'Distance']
columns.extend(f'{s}{lbl}' for lbl in ('Max', 'Min') for s in sigList )
df = pd.DataFrame([[a,b,*c,*d] for a,b,c,d in series.values], columns=columns)
print(df)
Length Distance sig1Max sig2Max sig3Max sig1Min sig2Min sig3Min
0 1 0 0.2 0.2 0.2 0.2 0.2 0.2
1 2 1000 0.6 0.7 0.5 0.1 0.3 0.1
2 1 0 0.4 0.4 0.4 0.4 0.4 0.4
3 1 0 0.5 0.5 0.5 0.5 0.5 0.5
4 3 14000 0.8 0.8 0.8 0.6 0.6 0.6
answered Nov 20 '18 at 23:26
coldspeedcoldspeed
135k23145231
edited Nov 21 '18 at 3:52
answered Nov 20 '18 at 23:26
coldspeedcoldspeed
135k23145231
answered Nov 20 '18 at 23:26
coldspeedcoldspeed
135k23145231
answered Nov 20 '18 at 23:26
coldspeedcoldspeed
135k23145231
135k23145231
Great, especially the dynamic one, since in my case sigList is resolved at runtime.
– burcak
Nov 21 '18 at 3:48
@burcak I had the order of the loops interchanged with the second one (sorry!) but you got the idea.
– coldspeed
Nov 21 '18 at 3:53
Thanks. In fact, the function below generates the series of tuples when called in apply method of grouped dataframe. Is there a way to just return dataframe so I won't need to create dataframe from series later on? def myfunc(x,signatures): return x.shape[0], x['start'].iloc[-1] - x['start'].iloc[0], x[signatures].agg(max), x[signatures].agg(min)
– burcak
Nov 21 '18 at 4:08
add a comment |
Great, especially the dynamic one, since in my case sigList is resolved at runtime.
– burcak
Nov 21 '18 at 3:48
@burcak I had the order of the loops interchanged with the second one (sorry!) but you got the idea.
– coldspeed
Nov 21 '18 at 3:53
Thanks. In fact, the function below generates the series of tuples when called in apply method of grouped dataframe. Is there a way to just return dataframe so I won't need to create dataframe from series later on? def myfunc(x,signatures): return x.shape[0], x['start'].iloc[-1] - x['start'].iloc[0], x[signatures].agg(max), x[signatures].agg(min)
– burcak
Nov 21 '18 at 4:08
Great, especially the dynamic one, since in my case sigList is resolved at runtime.
– burcak
Nov 21 '18 at 3:48
Great, especially the dynamic one, since in my case sigList is resolved at runtime.
– burcak
Nov 21 '18 at 3:48
@burcak I had the order of the loops interchanged with the second one (sorry!) but you got the idea.
– coldspeed
Nov 21 '18 at 3:53
@burcak I had the order of the loops interchanged with the second one (sorry!) but you got the idea.
– coldspeed
Nov 21 '18 at 3:53
Thanks. In fact, the function below generates the series of tuples when called in apply method of grouped dataframe. Is there a way to just return dataframe so I won't need to create dataframe from series later on? def myfunc(x,signatures): return x.shape[0], x['start'].iloc[-1] - x['start'].iloc[0], x[signatures].agg(max), x[signatures].agg(min)
– burcak
Nov 21 '18 at 4:08
Thanks. In fact, the function below generates the series of tuples when called in apply method of grouped dataframe. Is there a way to just return dataframe so I won't need to create dataframe from series later on? def myfunc(x,signatures): return x.shape[0], x['start'].iloc[-1] - x['start'].iloc[0], x[signatures].agg(max), x[signatures].agg(min)
– burcak
Nov 21 '18 at 4:08
add a comment |
1
You may check
newdf=pd.DataFrame(s.tolist())
newdf=pd.concat([newdf[[0,1]],pd.DataFrame(newdf[2].tolist()),pd.DataFrame(newdf[3].tolist())],1)
newdf.columns = [
"Length", "Distance", "sig1Max", "sig2Max", "sig3Max", "sig1Min", "sig2Min", "sig3Min"
]
newdf
Out[163]:
Length Distance sig1Max ... sig1Min sig2Min sig3Min
0 1 0 0.2 ... 0.2 0.2 0.2
1 2 1000 0.6 ... 0.1 0.3 0.1
2 1 0 0.4 ... 0.4 0.4 0.4
3 1 0 0.5 ... 0.5 0.5 0.5
4 3 14000 0.8 ... 0.6 0.6 0.6
[5 rows x 8 columns]
answered Nov 20 '18 at 23:17
Wen-BenWen-Ben
116k83369
Found a way as well!
– coldspeed
Nov 20 '18 at 23:26
add a comment |
1
You may check
newdf=pd.DataFrame(s.tolist())
newdf=pd.concat([newdf[[0,1]],pd.DataFrame(newdf[2].tolist()),pd.DataFrame(newdf[3].tolist())],1)
newdf.columns = [
"Length", "Distance", "sig1Max", "sig2Max", "sig3Max", "sig1Min", "sig2Min", "sig3Min"
]
newdf
Out[163]:
Length Distance sig1Max ... sig1Min sig2Min sig3Min
0 1 0 0.2 ... 0.2 0.2 0.2
1 2 1000 0.6 ... 0.1 0.3 0.1
2 1 0 0.4 ... 0.4 0.4 0.4
3 1 0 0.5 ... 0.5 0.5 0.5
4 3 14000 0.8 ... 0.6 0.6 0.6
[5 rows x 8 columns]
answered Nov 20 '18 at 23:17
Wen-BenWen-Ben
116k83369
Found a way as well!
– coldspeed
Nov 20 '18 at 23:26
add a comment |
1
1
1
You may check
newdf=pd.DataFrame(s.tolist())
newdf=pd.concat([newdf[[0,1]],pd.DataFrame(newdf[2].tolist()),pd.DataFrame(newdf[3].tolist())],1)
newdf.columns = [
"Length", "Distance", "sig1Max", "sig2Max", "sig3Max", "sig1Min", "sig2Min", "sig3Min"
]
newdf
Out[163]:
Length Distance sig1Max ... sig1Min sig2Min sig3Min
0 1 0 0.2 ... 0.2 0.2 0.2
1 2 1000 0.6 ... 0.1 0.3 0.1
2 1 0 0.4 ... 0.4 0.4 0.4
3 1 0 0.5 ... 0.5 0.5 0.5
4 3 14000 0.8 ... 0.6 0.6 0.6
[5 rows x 8 columns]
answered Nov 20 '18 at 23:17
Wen-BenWen-Ben
116k83369
You may check
newdf=pd.DataFrame(s.tolist())
newdf=pd.concat([newdf[[0,1]],pd.DataFrame(newdf[2].tolist()),pd.DataFrame(newdf[3].tolist())],1)
newdf.columns = [
"Length", "Distance", "sig1Max", "sig2Max", "sig3Max", "sig1Min", "sig2Min", "sig3Min"
]
newdf
Out[163]:
Length Distance sig1Max ... sig1Min sig2Min sig3Min
0 1 0 0.2 ... 0.2 0.2 0.2
1 2 1000 0.6 ... 0.1 0.3 0.1
2 1 0 0.4 ... 0.4 0.4 0.4
3 1 0 0.5 ... 0.5 0.5 0.5
4 3 14000 0.8 ... 0.6 0.6 0.6
[5 rows x 8 columns]
answered Nov 20 '18 at 23:17
Wen-BenWen-Ben
116k83369
answered Nov 20 '18 at 23:17
Wen-BenWen-Ben
116k83369
answered Nov 20 '18 at 23:17
Wen-BenWen-Ben
116k83369
answered Nov 20 '18 at 23:17
Wen-BenWen-Ben
116k83369
116k83369
Found a way as well!
– coldspeed
Nov 20 '18 at 23:26
add a comment |
Found a way as well!
– coldspeed
Nov 20 '18 at 23:26
Found a way as well!
– coldspeed
Nov 20 '18 at 23:26
Found a way as well!
– coldspeed
Nov 20 '18 at 23:26
add a comment |
0
You can flatten each element and then convert each to a Series itself. Converting each element to a Series turns the main Series (s in the example below) into a DataFrame. Then just set the column names as you wish.
For example:
import pandas as pd
# load in your data
s = pd.Series([
(1, 0, [0.2, 0.2, 0.2], [0.2, 0.2, 0.2]),
(2, 1000, [0.6, 0.7, 0.5], [0.1, 0.3, 0.1]),
(1, 0, [0.4, 0.4, 0.4], [0.4, 0.4, 0.4]),
(1, 0, [0.5, 0.5, 0.5], [0.5, 0.5, 0.5]),
(3, 14000, [0.8, 0.8, 0.8], [0.6, 0.6, 0.6]),
])
def flatten(x):
# note this is not very robust, but works for this case
return [x[0], x[1], *x[2], *x[3]]
df = s.apply(flatten).apply(pd.Series)
df.columns = [
"Length", "Distance", "sig1Max", "sig2Max", "sig3Max", "sig1Min", "sig2Min", "sig3Min"
]
Then you have df as:
Length Distance sig1Max sig2Max sig3Max sig1Min sig2Min sig3Min
0 1.0 0.0 0.2 0.2 0.2 0.2 0.2 0.2
1 2.0 1000.0 0.6 0.7 0.5 0.1 0.3 0.1
2 1.0 0.0 0.4 0.4 0.4 0.4 0.4 0.4
3 1.0 0.0 0.5 0.5 0.5 0.5 0.5 0.5
4 3.0 14000.0 0.8 0.8 0.8 0.6 0.6 0.6
answered Nov 20 '18 at 22:41
Henry WoodyHenry Woody
4,7273927
add a comment |
0
You can flatten each element and then convert each to a Series itself. Converting each element to a Series turns the main Series (s in the example below) into a DataFrame. Then just set the column names as you wish.
For example:
import pandas as pd
# load in your data
s = pd.Series([
(1, 0, [0.2, 0.2, 0.2], [0.2, 0.2, 0.2]),
(2, 1000, [0.6, 0.7, 0.5], [0.1, 0.3, 0.1]),
(1, 0, [0.4, 0.4, 0.4], [0.4, 0.4, 0.4]),
(1, 0, [0.5, 0.5, 0.5], [0.5, 0.5, 0.5]),
(3, 14000, [0.8, 0.8, 0.8], [0.6, 0.6, 0.6]),
])
def flatten(x):
# note this is not very robust, but works for this case
return [x[0], x[1], *x[2], *x[3]]
df = s.apply(flatten).apply(pd.Series)
df.columns = [
"Length", "Distance", "sig1Max", "sig2Max", "sig3Max", "sig1Min", "sig2Min", "sig3Min"
]
Then you have df as:
Length Distance sig1Max sig2Max sig3Max sig1Min sig2Min sig3Min
0 1.0 0.0 0.2 0.2 0.2 0.2 0.2 0.2
1 2.0 1000.0 0.6 0.7 0.5 0.1 0.3 0.1
2 1.0 0.0 0.4 0.4 0.4 0.4 0.4 0.4
3 1.0 0.0 0.5 0.5 0.5 0.5 0.5 0.5
4 3.0 14000.0 0.8 0.8 0.8 0.6 0.6 0.6
answered Nov 20 '18 at 22:41
Henry WoodyHenry Woody
4,7273927
add a comment |
0
0
0
You can flatten each element and then convert each to a Series itself. Converting each element to a Series turns the main Series (s in the example below) into a DataFrame. Then just set the column names as you wish.
For example:
import pandas as pd
# load in your data
s = pd.Series([
(1, 0, [0.2, 0.2, 0.2], [0.2, 0.2, 0.2]),
(2, 1000, [0.6, 0.7, 0.5], [0.1, 0.3, 0.1]),
(1, 0, [0.4, 0.4, 0.4], [0.4, 0.4, 0.4]),
(1, 0, [0.5, 0.5, 0.5], [0.5, 0.5, 0.5]),
(3, 14000, [0.8, 0.8, 0.8], [0.6, 0.6, 0.6]),
])
def flatten(x):
# note this is not very robust, but works for this case
return [x[0], x[1], *x[2], *x[3]]
df = s.apply(flatten).apply(pd.Series)
df.columns = [
"Length", "Distance", "sig1Max", "sig2Max", "sig3Max", "sig1Min", "sig2Min", "sig3Min"
]
Then you have df as:
Length Distance sig1Max sig2Max sig3Max sig1Min sig2Min sig3Min
0 1.0 0.0 0.2 0.2 0.2 0.2 0.2 0.2
1 2.0 1000.0 0.6 0.7 0.5 0.1 0.3 0.1
2 1.0 0.0 0.4 0.4 0.4 0.4 0.4 0.4
3 1.0 0.0 0.5 0.5 0.5 0.5 0.5 0.5
4 3.0 14000.0 0.8 0.8 0.8 0.6 0.6 0.6
answered Nov 20 '18 at 22:41
Henry WoodyHenry Woody
4,7273927
You can flatten each element and then convert each to a Series itself. Converting each element to a Series turns the main Series (s in the example below) into a DataFrame. Then just set the column names as you wish.
For example:
import pandas as pd
# load in your data
s = pd.Series([
(1, 0, [0.2, 0.2, 0.2], [0.2, 0.2, 0.2]),
(2, 1000, [0.6, 0.7, 0.5], [0.1, 0.3, 0.1]),
(1, 0, [0.4, 0.4, 0.4], [0.4, 0.4, 0.4]),
(1, 0, [0.5, 0.5, 0.5], [0.5, 0.5, 0.5]),
(3, 14000, [0.8, 0.8, 0.8], [0.6, 0.6, 0.6]),
])
def flatten(x):
# note this is not very robust, but works for this case
return [x[0], x[1], *x[2], *x[3]]
df = s.apply(flatten).apply(pd.Series)
df.columns = [
"Length", "Distance", "sig1Max", "sig2Max", "sig3Max", "sig1Min", "sig2Min", "sig3Min"
]
Then you have df as:
Length Distance sig1Max sig2Max sig3Max sig1Min sig2Min sig3Min
0 1.0 0.0 0.2 0.2 0.2 0.2 0.2 0.2
1 2.0 1000.0 0.6 0.7 0.5 0.1 0.3 0.1
2 1.0 0.0 0.4 0.4 0.4 0.4 0.4 0.4
3 1.0 0.0 0.5 0.5 0.5 0.5 0.5 0.5
4 3.0 14000.0 0.8 0.8 0.8 0.6 0.6 0.6
answered Nov 20 '18 at 22:41
Henry WoodyHenry Woody
4,7273927
answered Nov 20 '18 at 22:41
Henry WoodyHenry Woody
4,7273927
answered Nov 20 '18 at 22:41
Henry WoodyHenry Woody
4,7273927
answered Nov 20 '18 at 22:41
Henry WoodyHenry Woody
4,7273927
4,7273927
add a comment |
add a comment |
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def myfunc(x,signatures): return x.shape[0], x['start'].iloc[-1] - x['start'].iloc[0], x[signatures].agg(max), x[signatures].agg(min)This function creates the lists in tuples. Can we flatten the list in the return statement.– burcak
Nov 20 '18 at 22:59