I'm starting with a dataframe of baseabll seasons a section of which looks similar to this:
Name Season AB H SB playerid
13047 A.J. Pierzynski 2013 503 137 1 746
6891 A.J. Pierzynski 2006 509 150 1 746
1374 Rod Carew 1977 616 239 23 1001942
1422 Stan Musial 1948 611 230 7 1009405
1507 Todd Helton 2000 580 216 5 432
1508 Nomar Garciaparra 2000 529 197 5 190
1509 Ichiro Suzuki 2004 704 262 36 1101
From these seasons, I want to create a dataframe of career stats; that is, one row for each player which is a sum of their AB, H, etc. This dataframe should still include the names of the players. The playerid in the above is a unique key for each player and should either be an index or an unchanged value in a column after creating the career stats dataframe.
My hypothetical starting point is df_careers = df_seasons.groupby('playerid').agg(sum) but this leaves out all the non-numeric data. With numeric_only = False I can get some sort of mess in the names columns like 'Ichiro SuzukiIchiro SuzukiIchiro Suzuki' from concatenation, but that just requires a bunch of cleaning. This is something I'd like to be able to do with other data sets and the actually data I have is more like 25 columns, so I'd rather understand a specific routine for getting the Name data back or preserving it from the outset rather than write a specific function and use groupby('playerid').agg(func) (or a similar process) to do it, if possible.
I'm guessing there's a fairly simply way to do this, but I only started learning Pandas a week ago, so there are gaps in my knowledge.
You can write your own condition how do you want to include non summed columns.
col = df.columns.tolist()
col.remove('playerid')
df.groupby('playerid').agg({i : lambda x: x.iloc[0] if x.dtypes=='object' else x.sum() for i in df.columns})
df:
Name Season AB H SB playerid
playerid
190 Nomar_Garciaparra 2000 529 197 5 190
432 Todd_Helton 2000 580 216 5 432
746 A.J._Pierzynski 4019 1012 287 2 1492
1101 Ichiro_Suzuki 2004 704 262 36 1101
1001942 Rod_Carew 1977 616 239 23 1001942
1009405 Stan_Musial 1948 611 230 7 1009405
If there is a one-to-one relationship between 'playerid' and 'Name', as appears to be the case, you can just include 'Name' in the groupby columns:
stat_cols = ['AB', 'H', 'SB']
groupby_cols = ['playerid', 'Name']
results = df.groupby(groupby_cols)[stat_cols].sum()
Results:
AB H SB
playerid Name
190 Nomar Garciaparra 529 197 5
432 Todd Helton 580 216 5
746 A.J. Pierzynski 1012 287 2
1101 Ichiro Suzuki 704 262 36
1001942 Rod Carew 616 239 23
1009405 Stan Musial 611 230 7
If you'd prefer to group only by 'playerid' and add the 'Name' data back in afterwards, you can instead create a 'playerId' to 'Name' mapping as a dictionary, and look it up using map:
results = df.groupby('playerid')[stat_cols].sum()
name_map = pd.Series(df.Name.to_numpy(), df.playerid).to_dict()
results['Name'] = results.index.map(name_map)
Results:
AB H SB Name
playerid
190 529 197 5 Nomar Garciaparra
432 580 216 5 Todd Helton
746 1012 287 2 A.J. Pierzynski
1101 704 262 36 Ichiro Suzuki
1001942 616 239 23 Rod Carew
1009405 611 230 7 Stan Musial
groupy.agg() can accept a dictionary that maps column names to functions. So, one solution is to pass a dictionary to agg, specifying which functions to apply to each column.
Using the sample data above, one might use
mapping = { 'AB': sum,'H': sum, 'SB': sum, 'Season': max, 'Name': max }
df_1 = df.groupby('playerid').agg(mapping)
The choice to use 'max' for those that shouldn't be summed is arbitrary. You could define a lambda function to apply to a column if you want to handle it in a certain way. DataFrameGroupBy.agg can work with any function that will work with DataFrame.apply.
To expand this to larger data sets, you might use a dictionary comprehension. This would work well:
dictionary = { x : sum for x in df.columns}
dont_sum = {'Name': max, 'Season': max}
dictionary.update(dont_sum)
df_1 = df.groupby('playerid').agg(dictionary)
I am looking for a way to aggregate over a large dataframe, possibly using groupby. Each group would be based on either pre-specified columns or regex, and the aggregation should produce a named output.
This produces a sample dataframe:
import pandas as pd
import itertools
import numpy as np
col = "A,B,C".split(',')
col1 = "1,2,3,4,5,6,7,8,9".split(',')
col2 = "E,F,G".split(',')
all_dims = [col, col1, col2]
all_keys = ['.'.join(i) for i in itertools.product(*all_dims)]
rng = pd.date_range(end=pd.Timestamp.today().date(), periods=12, freq='M')
df = pd.DataFrame(np.random.randint(0, 1000, size=(len(rng), len(all_keys))), columns=all_keys, index=rng)
Above produces a dataframe with one year's worth of monthly data, with 36 columns with following names:
['A.1.E', 'A.1.F', 'A.1.G', 'A.2.E', 'A.2.F', 'A.2.G', 'A.3.E', 'A.3.F',
'A.3.G', 'A.4.E', 'A.4.F', 'A.4.G', 'A.5.E', 'A.5.F', 'A.5.G', 'A.6.E',
'A.6.F', 'A.6.G', 'A.7.E', 'A.7.F', 'A.7.G', 'A.8.E', 'A.8.F', 'A.8.G',
'A.9.E', 'A.9.F', 'A.9.G', 'B.1.E', 'B.1.F', 'B.1.G', 'B.2.E', 'B.2.F',
'B.2.G', 'B.3.E', 'B.3.F', 'B.3.G', 'B.4.E', 'B.4.F', 'B.4.G', 'B.5.E',
'B.5.F', 'B.5.G', 'B.6.E', 'B.6.F', 'B.6.G', 'B.7.E', 'B.7.F', 'B.7.G',
'B.8.E', 'B.8.F', 'B.8.G', 'B.9.E', 'B.9.F', 'B.9.G', 'C.1.E', 'C.1.F',
'C.1.G', 'C.2.E', 'C.2.F', 'C.2.G', 'C.3.E', 'C.3.F', 'C.3.G', 'C.4.E',
'C.4.F', 'C.4.G', 'C.5.E', 'C.5.F', 'C.5.G', 'C.6.E', 'C.6.F', 'C.6.G',
'C.7.E', 'C.7.F', 'C.7.G', 'C.8.E', 'C.8.F', 'C.8.G', 'C.9.E', 'C.9.F',
'C.9.G']
What I would like now is to be able aggregate over the dataframe and take certain column combinations and produce named outputs. For example, one rules might be that I will take all 'A.*.E' columns (that have any number in the middle), sum them and produce a named output column called 'A.SUM.E'. And then do the same for 'A.*.F', 'A.*.G' and so on.
I have looked into pandas 25 named aggregation which allows me to name my outputs but I couldn't see how to simultaneously capture the right column combinations and produce the right output names.
If you need to reshape the dataframe to make a workable solution, that is fine as well.
Note, I am aware I could do something like this in a Python loop but I am looking for a pandas way to do it.
Not a groupby solution and it uses a loop but I think it's nontheless rather elegant: first get a list of unique column from - to combinations using a set and then do the sums using filter:
cols = sorted([(x[0],x[1]) for x in set([(x.split('.')[0], x.split('.')[-1]) for x in df.columns])])
for c0, c1 in cols:
df[f'{c0}.SUM.{c1}'] = df.filter(regex = f'{c0}\.\d+\.{c1}').sum(axis=1)
Result:
A.1.E A.1.F A.1.G A.2.E ... B.SUM.G C.SUM.E C.SUM.F C.SUM.G
2018-08-31 978 746 408 109 ... 4061 5413 4102 4908
2018-09-30 923 649 488 447 ... 5585 3634 3857 4228
2018-10-31 911 359 897 425 ... 5039 2961 5246 4126
2018-11-30 77 479 536 509 ... 4634 4325 2975 4249
2018-12-31 608 995 114 603 ... 5377 5277 4509 3499
2019-01-31 138 612 363 218 ... 4514 5088 4599 4835
2019-02-28 994 148 933 990 ... 3907 4310 3906 3552
2019-03-31 950 931 209 915 ... 4354 5877 4677 5557
2019-04-30 255 168 357 800 ... 5267 5200 3689 5001
2019-05-31 593 594 824 986 ... 4221 2108 4636 3606
2019-06-30 975 396 919 242 ... 3841 4787 4556 3141
2019-07-31 350 312 104 113 ... 4071 5073 4829 3717
If you want to have the result in a new DataFrame, just create an empty one and add the columns to it:
result = pd.DataFrame()
for c0, c1 in cols:
result[f'{c0}.SUM.{c1}'] = df.filter(regex = f'{c0}\.\d+\.{c1}').sum(axis=1)
Update: using simple groupby (which is even more simple in this particular case):
def grouper(col):
c = col.split('.')
return f'{c[0]}.SUM.{c[-1]}'
df.groupby(grouper, axis=1).sum()
I have a variable in the following format fg = 2017-20. It's a string. And also I have a dataframe:
flag №
2017-18 389
2017-19 390
2017-20 391
2017-21 392
2017-22 393
2017-23 394
...
I need to find this value (fg) in the column "flag" and select the appropriate value (in the example it will be 391) in the column "№". Then create new dataframe, in which there will also be a column "№". Add this value to this dataframe and iterate 53 times. The result should look like this:
№_new
391
392
393
394
395
...
442
443
444
It does not look difficult, but I can not find anything suitable based on other issues. Can someone advise anything, please?
You need boolean indexing with loc for filtering, then convert one item Series to scalar by convert to numpy array by values and select first value by [0].
Last create new DataFrame with numpy.arange.
fg = '2017-20'
val = df.loc[df['flag'] == fg, '№'].values[0]
print (val)
391
df1 = pd.DataFrame({'№_new':np.arange(val, val+53)})
print (df1)
№_new
0 391
1 392
2 393
3 394
4 395
5 396
6 397
7 398
8 399
9 400
10 401
11 402
..
..
Input:
## x2
##0 214
##1 234
##2 253
##3 272
##4 291
Desired output:
## x2
##0 291
##1 214
##2 234
##3 253
##4 272
Following code eliminates bottom part of the shifted column and puts NA on top. However, I want it as a cycle.
a = pd.DataFrame([214,234,253,272,291], columns=['x2'])
a.x2 = a.x2.shift(1)
I would just chain a call to fill fillna after the call to shift:
import pandas
a = pandas.DataFrame([214,234,253,272,291], columns=['x2'])
a['x3'] = a.shift(1).fillna(a['x2'].iloc[-1])
print(a)
x2 x3
0 214 291
1 234 214
2 253 234
3 272 253
4 291 272
You can reassign directly to the x2 column, but I wanted to be able to show both the source and the result a once for comparison
I have a data frame that is structured similar to the following (but in the real case with many more rows and columns).
In [2]: Ex # The example DataFrame
Out[2]:
NameBef v1B v2B v3B v4B NameAft v1A v2A v3A v4A
Id
422 firstBef 133 145 534 745 FirstAft 212 543 2342 4563
862 secondBef 234 434 345 3453 SecondAft 643 493 3433 234
935 thirdBef 232 343 6454 463 thirdAft 423 753 754 743
For each row I want to calculate the quotient each vXB and vXA value from above (the Xs are variables) to end up with a DataFrame like this one
v1Q v2Q v3Q v4Q
Id
422 1.593985 3.744828 4.385768 6.124832
862 2.747863 1.135945 9.950725 0.067767
935 1.823276 2.195335 0.116827 1.604752
Where each element is the quotient of the corresponding elements of the original data frame.
I haven't been able to figure out how to do this conveniently.
To be convenient it would be good if it will not be required to provide only the names of the first and last columns of the "before" and "after" values i.e. 'v1B', 'v4B' and 'v1A', 'v4A' (i.e. not each of the columns).
The following is what I have come up with.
In [3]: C=Ex.columns
In [4]: C1B=C.get_loc('v1B')
In [5]: C2B=C.get_loc('v4B')
In [6]: C1A=C.get_loc('v1A')
In [7]: C2A=C.get_loc('v4A')
In [8]: FB=Ex.ix[:,C1B:C2B+1]
In [9]: FA=Ex.ix[:,C1A:C2A+1]
In [10]: FB # The FB and FA frames have this structure
Out[10]:
v1B v2B v3B v4B
Id
422 133 145 534 745
862 234 434 345 3453
935 232 343 6454 463
[3 rows x 4 columns]
Then finally produce the required DataFrame. This is done by doing the calculation on numpy arrays produced by DataFrame.values.
This method I got from this question/answer:
In [12]: DataFrame((FA.values*1.0) / FB.values,columns=['v1Q','v2Q','v3Q','v4Q'],index=Ex.index)
Out[12]:
v1Q v2Q v3Q v4Q
Id
422 1.593985 3.744828 4.385768 6.124832
862 2.747863 1.135945 9.950725 0.067767
935 1.823276 2.195335 0.116827 1.604752
[3 rows x 4 columns]
Am I missing something? I was hoping that I could achieve this in some much more direct way by doing some operation on the original DataFrame.
Is there no operation to do elementwise calculation directly on DataFrames instead of going via numpy arrays?
You could always use df.filter to select the relevant column names. It can accept a regular expression so you could specify the after/before columns with something like this:
>>> df.filter(regex=r'^v.A$').values / df.filter(regex=r'^v.B$').values
array([[ 1.59398496, 3.74482759, 4.38576779, 6.12483221],
[ 2.74786325, 1.1359447 , 9.95072464, 0.06776716],
[ 1.82327586, 2.19533528, 0.11682677, 1.60475162]])
Regarding the arithmetic operation, you're not missing anything. It's necessary to use Numpy arrays (.values) here as otherwise Pandas computes values from the common index labels in both DataFrames. If an index is missing the calculation results in NaN. Numpy arrays don't have labeled indexes so the element-wise operation succeeds.