We Keep Coding

Python, lambda function as argument for groupby

I'm trying to figure out what a piece of code is doing, but I'm getting kinda lost on it.
I have a pandas dataframe, which has been loaded by the following .csv file:
origin_census_block_group,date_range_start,date_range_end,device_count,distance_traveled_from_home,bucketed_distance_traveled,median_dwell_at_bucketed_distance_traveled,completely_home_device_count,median_home_dwell_time,bucketed_home_dwell_time,at_home_by_each_hour,part_time_work_behavior_devices,full_time_work_behavior_devices,destination_cbgs,delivery_behavior_devices,median_non_home_dwell_time,candidate_device_count,bucketed_away_from_home_time,median_percentage_time_home,bucketed_percentage_time_home,mean_home_dwell_time,mean_non_home_dwell_time,mean_distance_traveled_from_home
010539707003,2020-06-25T00:00:00-05:00,2020-06-26T00:00:00-05:00,49,626,"{""16001-50000"":5,""0"":11,"">50000"":4,""2001-8000"":3,""1-1000"":9,""1001-2000"":7,""8001-16000"":1}","{""16001-50000"":110,"">50000"":155,""<1000"":40,""2001-8000"":237,""1001-2000"":27,""8001-16000"":180}",12,627,"{""721-1080"":11,""361-720"":9,""61-360"":1,""<60"":11,"">1080"":12}","[32,32,28,30,30,31,27,23,20,20,20,17,19,19,15,14,17,20,20,21,25,22,24,23]",7,3,"{""120330012011"":1,""010030107031"":1,""010030114052"":2,""120330038001"":1,""010539701003"":1,""010030108001"":1,""010539707002"":14,""010539705003"":2,""120330015001"":1,""121130102003"":1,""010539701002"":1,""120330040001"":1,""370350101014"":2,""120330033081"":2,""010030106003"":1,""010539706001"":2,""010539707004"":3,""120330039001"":1,""010539699003"":1,""120330030003"":1,""010539707003"":41,""010970029003"":1,""010539705004"":1,""120330009002"":1,""010539705001"":3,""010539704003"":1,""120330028012"":1,""120330035081"":1,""120330036102"":1,""120330036142"":1,""010030114062"":1,""010539706004"":7,""010539706002"":1,""120330036082"":1,""010539707001"":7,""010030102001"":1,""120330028011"":1}",2,241,71,"{""21-45"":4,""481-540"":2,""541-600"":1,""721-840"":1,""1201-1320"":1,""301-360"":3,""<20"":13,""61-120"":3,""241-300"":3,""121-180"":1,""421-480"":3,""1321-1440"":4,""1081-1200"":1,""961-1080"":2,""601-660"":1,""181-240"":1,""661-720"":2,""361-420"":3}",72,"{""0-25"":13,""76-100"":21,""51-75"":6,""26-50"":3}",657,413,1936
010730144081,2020-06-25T00:00:00-05:00,2020-06-26T00:00:00-05:00,139,2211,"{""16001-50000"":17,""0"":41,"">50000"":15,""2001-8000"":22,""1-1000"":8,""1001-2000"":12,""8001-16000"":24}","{""16001-50000"":143,"">50000"":104,""<1000"":132,""2001-8000"":39,""1001-2000"":15,""8001-16000"":102}",41,806,"{""721-1080"":32,""361-720"":16,""61-360"":12,""<60"":30,"">1080"":46}","[91,92,93,91,91,90,86,83,78,64,64,61,64,62,65,62,60,74,61,64,75,78,81,84]",8,6,"{""131350501064"":1,""131350502151"":1,""010730102002"":1,""011170302131"":2,""010730038024"":1,""010730108041"":1,""010730144133"":1,""010730132003"":1,""011210118002"":1,""011170303053"":1,""010730111084"":2,""011170302142"":1,""010730119011"":1,""010730129063"":2,""010730107063"":1,""010730059083"":1,""010730058003"":1,""011270204003"":1,""010730049012"":2,""130879701001"":1,""010730120021"":1,""130890219133"":1,""010730144082"":4,""170310301031"":1,""010730129112"":1,""010730024002"":1,""011170303034"":2,""481390616004"":1,""121270826052"":1,""010730128021"":2,""121270825073"":1,""010730004004"":1,""211959313002"":1,""010730100012"":1,""011170302151"":1,""010730142041"":1,""010730129123"":1,""010730129084"":1,""010730042002"":1,""010730059033"":2,""170318306001"":1,""130519800001"":1,""010730027003"":1,""121270826042"":1,""481610001002"":1,""010730100011"":1,""010730023032"":1,""350250004002"":1,""010730056003"":1,""010730132001"":1,""011170302171"":2,""120910227003"":1,""011239620001"":1,""130351503002"":1,""010730129155"":1,""010730001001"":2,""010730110021"":1,""170310104003"":1,""010730059082"":2,""010730120022"":1,""011170303151"":1,""010730139022"":1,""011170303441"":4,""010730144092"":3,""010730129151"":1,""011210119001"":2,""010730144081"":117,""010730108052"":1,""010730129122"":9,""370710321003"":1,""010730142034"":2,""010730042001"":2,""010570201003"":1,""010730144132"":6,""010730059032"":1,""010730012001"":2,""010730102003"":1,""011170303332"":1,""010730128032"":2,""010730129081"":1,""010730103011"":1,""010730058001"":3,""011150401041"":1,""010730045001"":3,""010730110013"":1,""010730119041"":1,""010730042003"":1,""010730141041"":1,""010730144091"":1,""010730129154"":1,""484759501002"":1,""010730144063"":1,""010730144102"":12,""011170303141"":1,""011250106011"":1,""011170303152"":1,""010730059104"":1,""010730107021"":1,""010730100014"":1,""010730008004"":1,""011170303451"":1,""010730127041"":2,""370559704003"":1,""010730047011"":2,""010730129132"":2,""011010014002"":1,""010730144131"":1,""011170302133"":1,""010730030011"":1,""131350506063"":1,""010730118023"":1,""010890110141"":1,""010730128023"":1,""010730106022"":2,""130879703004"":1,""010730108015"":1,""131390010041"":1,""011170305013"":1,""010730134002"":1,""010730031004"":1,""010730138012"":1,""010730011004"":1,""011250102041"":1,""010730129131"":4,""010730144101"":4,""011170303331"":2,""010730003001"":1,""011010033012"":1,""483539504004"":1,""010550104021"":1,""011170303411"":1,""010730106031"":1,""011170303153"":5,""010730128034"":1,""010730129061"":1,""131390010023"":1,""010730051042"":1,""130510107002"":1,""010730027001"":2,""120090686011"":1,""010730107042"":1,""010730123052"":1,""010730129102"":1,""011210115003"":1,""010730129083"":4,""011170303142"":1,""011010014001"":1,""010730107064"":2}",7,176,205,"{""21-45"":7,""481-540"":10,""541-600"":4,""46-60"":2,""721-840"":3,""1201-1320"":3,""301-360"":7,""<20"":46,""61-120"":6,""241-300"":4,""121-180"":9,""421-480"":2,""1321-1440"":3,""1081-1200"":5,""961-1080"":1,""601-660"":1,""181-240"":5,""661-720"":1,""361-420"":7}",78,"{""0-25"":29,""76-100"":71,""51-75"":27,""26-50"":8}",751,338,38937
010890017002,2020-06-25T00:00:00-05:00,2020-06-26T00:00:00-05:00,78,1934,"{""16001-50000"":2,""0"":12,"">50000"":9,""2001-8000"":27,""1-1000"":12,""1001-2000"":8,""8001-16000"":8}","{""16001-50000"":49,"">50000"":99,""<1000"":111,""2001-8000"":37,""1001-2000"":24,""8001-16000"":28}",11,787,"{""721-1080"":17,""361-720"":11,""61-360"":11,""<60"":15,"">1080"":23}","[49,42,48,48,47,48,44,44,39,32,34,32,36,31,32,36,40,37,36,38,49,45,46,46]",5,1,"{""010890101002"":1,""010730108041"":1,""010890020003"":2,""010890010001"":2,""010890025011"":3,""010890026001"":4,""280819505003"":1,""281059504004"":1,""010890103022"":1,""120990056011"":1,""010890109012"":2,""010890019021"":6,""010890013021"":4,""010890015004"":3,""010890108003"":1,""010890014022"":6,""281059501003"":1,""281059503001"":1,""010890007022"":3,""010890017001"":3,""010890107023"":1,""010890021002"":1,""010890009011"":1,""010890109013"":1,""010730120022"":1,""010890031003"":15,""011170303151"":1,""010890019011"":9,""010890030002"":2,""010890110221"":1,""011170305021"":1,""010890026003"":2,""010890025012"":3,""010730117034"":1,""010830208022"":1,""010890031002"":2,""010890112002"":1,""010210602001"":1,""010890002022"":1,""010890017002"":65,""281059506021"":1,""010890010003"":2,""010890106222"":1,""120990059182"":1,""010890110222"":1,""010890020001"":1,""010890101003"":1,""010890018013"":1,""010890021001"":1,""010890109021"":1,""010890108001"":1,""010770106005"":1,""281059506011"":1,""010030114032"":2,""010830209001"":1,""010890027222"":1,""010730128023"":1,""010890009021"":1,""010030114051"":1,""010030109031"":1,""010030103003"":1,""010890031001"":1,""010890021003"":1,""010030114062"":4,""010890106241"":1,""281059504003"":1,""010890018011"":10,""010890019031"":5,""010890027012"":1,""010730108054"":1,""010890106223"":2,""010890111001"":1,""010210603002"":1,""010890109011"":1,""010890019012"":2,""010890113001"":1,""010890028013"":3}",1,229,99,"{""481-540"":3,""541-600"":2,""46-60"":1,""721-840"":1,""1201-1320"":7,""301-360"":6,""<20"":18,""61-120"":10,""241-300"":5,""121-180"":2,""1321-1440"":2,""841-960"":1,""1081-1200"":1,""961-1080"":3,""601-660"":3,""181-240"":2,""661-720"":3}",78,"{""0-25"":16,""76-100"":44,""51-75"":11,""26-50"":7}",708,353,14328
010950308022,2020-06-25T00:00:00-05:00,2020-06-26T00:00:00-05:00,100,2481,"{""16001-50000"":11,""0"":19,"">50000"":11,""2001-8000"":40,""1-1000"":6,""1001-2000"":3,""8001-16000"":4}","{""16001-50000"":150,"">50000"":23,""<1000"":739,""2001-8000"":23,""1001-2000"":12,""8001-16000"":208}",17,703,"{""721-1080"":21,""361-720"":19,""61-360"":10,""<60"":24,"">1080"":26}","[62,64,64,63,65,67,54,48,37,37,34,33,30,34,32,33,35,43,50,56,58,56,56,57]",8,6,"{""010950306004"":1,""010950302023"":1,""011030054051"":1,""010950311002"":1,""010950309023"":1,""010499606003"":1,""121319506023"":2,""010950308022"":86,""121319506016"":2,""010950304013"":1,""010950307024"":1,""010950309041"":1,""010890019021"":2,""010950312001"":5,""010499607002"":1,""011150402013"":1,""010550102003"":1,""120050027043"":3,""010719509003"":1,""010950302022"":1,""010950308023"":2,""120050027051"":2,""471079701022"":1,""010890106221"":1,""010950306001"":1,""010950302011"":2,""011150405013"":1,""011150402041"":2,""010950312002"":16,""011030054042"":1,""010950301002"":2,""130459105011"":1,""010730001001"":1,""130459102001"":1,""010890109013"":2,""010950308013"":14,""010719508004"":1,""120050027041"":3,""010550110021"":3,""010730049022"":1,""010950308024"":1,""010950312004"":6,""010950312003"":1,""010550104012"":2,""010550110013"":1,""120860004111"":1,""010890027222"":1,""010950306002"":2,""010950304015"":1,""011030054041"":1,""010950309031"":8,""010950308021"":1,""010950302024"":1,""010950307011"":5,""010550110012"":2,""011150404013"":1,""130459103003"":1,""120050027032"":3,""010950307012"":5,""010950309022"":2,""010950307023"":1,""010719508003"":1,""010499608001"":2,""010950310003"":1,""011150402043"":1,""120860099063"":1,""010950309021"":4,""010950309043"":2,""010950308011"":1,""010950306003"":3,""120050027042"":1,""010950308025"":5,""010950309032"":6,""010499607001"":1}",1,199,132,"{""21-45"":8,""481-540"":6,""541-600"":4,""46-60"":3,""721-840"":3,""1201-1320"":4,""301-360"":3,""<20"":20,""61-120"":10,""241-300"":2,""121-180"":4,""421-480"":3,""1321-1440"":1,""841-960"":3,""961-1080"":2,""601-660"":1,""181-240"":3,""661-720"":1,""361-420"":2}",74,"{""0-25"":20,""76-100"":48,""51-75"":23,""26-50"":4}",661,350,5044
df = pd.read_csv(csv_file,
usecols=[
'origin_census_block_group',
'date_range_start',
'date_range_end',
'device_count',
'distance_traveled_from_home',
'completely_home_device_count',
'median_home_dwell_time',
'part_time_work_behavior_devices',
'full_time_work_behavior_devices'
],
dtype={'origin_census_block_group': str},
).set_index('origin_census_block_group')
and, later in the code, the dataframe is modified by:
df = df.groupby(lambda cbg: cbg[:5]).sum()
I don't quite understand what this line is doing precisely.
Groupby generally groups a dataframe by column, so...is it grouping the dataframe using multiple columns (0 to 5)? What is the effect of .sum() at the end?

If you run your code exactly as you wrote it (both the creation of df and the groupby) you can see the result. I print first couple of columns of the output of groupby
device_count distance_traveled_from_home
----- -------------- -----------------------------
01053 49 626
01073 139 2211
01089 78 1934
01095 100 2481
What happens here is the function lambda cbg: cbg[:5] is applied to each of the index values (strings that look like numbers in column origin_census_block_group). As a side, note the statement
...
dtype={'origin_census_block_group': str},
when creating the df, so somebody went into trouble to make sure they are actually str
So the function is applied to string like '010539707003' and returns a substring which is the first 5 characters of that string:
'010539707003'[:5]
produces
'01053'
so I assume there are multiple keys that share the first 5 characters (in the actual file -- the snippet has them all unique so not very interesting) and all these rows are grouped together
Then .sum() is applied to each numerical column of each group and returns, well, the column sum per each groupby key. This is what you see in my output in column 'device_count' and so on.
Hope this is clear now

Pandas' read_csv() will render a csv-formatted file a Pandas Dataframe.
I recommend having a ready at the Pandas' documentation, as it's very exhaustive -> https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.read_csv.html
usecols=[
'origin_census_block_group',
'date_range_start',
'date_range_end',
'device_count',
'distance_traveled_from_home',
'completely_home_device_count',
'median_home_dwell_time',
'part_time_work_behavior_devices',
'full_time_work_behavior_devices'
],
The usecols parameter will take as input an array of desired columns and will only load the specified columns into the dataframe.
dtype={'origin_census_block_group': str}
The dtype parameter will take a dict as input and is to specify the data type of the values, like {'column' : datatype}
.set_index('origin_census_block_group')
.set_index() will set the specificed column as the index column (ie: the first column). The usual index of Pandas' Dataframe is the row's index number, which appears as the first column of the dataframe. By setting the index, the first column now becomes the specified column. See: https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.set_index.html
Panda's .groupby() function will take a dataframe a regroup it basing on the occurrences of he values from the specified column.
That is to say, if we a dataframe such as df =
Fruit Name Quality Count
Apple Marco High 4
Pear Lucia Medium 10
Apple Francesco Low 3
Banana Carlo Medium 6
Pear Timmy Low 7
Apple Roberto High 8
Banana Joe High 21
Banana Jack Low 3
Pear Rob Medium 5
Apple Louis Medium 6
Pear Jennifer Low 7
Pear Laura High 8
Performing a groupby operations, such as:
df = df.groupby(lambda x: x[:2]).sum()
Will take all the elements in the index, slice them from index 0 through index 2 and return the sum of all the corresponding values, ie:
Ap 21
Ba 30
Pe 37
Now, you might be wondering about that final .sum() method. If you try to print the dataframe without applying it, you'll likely get something like this:
<bound method GroupBy.sum of <pandas.core.groupby.generic.DataFrameGroupBy object at 0x109d260a0>>
This is because Pandas has created a groubpy object and does not yet now how to display it to you. Do you want to have it displayed by the number of the occurrences in the index? You'd do this:
df = df.groupby(lambda x: x[:2]).size()
And that would output:
Ap 4
Ba 3
Pe 5
Or maybe the sum of their respective summable values? (Which is what is done in the example)
df = df.groupby(lambda x: x[:2]).sum()
Which again, will output:
Ap 21
Ba 30
Pe 37
Notice it has taken the first two letters of the string in the index. Had it been x[:3], it would have taken the first three letters, of course.
Summing it up:
-> .groupby() takes the elements in the index, i.e. the first column of the dataframe and organises the dataframe in groups relating to the index
-> The input you have given to groubpy is an anonymous function, i.e. lambda function, slicing from index 0 through 5 of its mapped input
-> You may choose how to have the results of groubpy by appending the methos .sum() or .size() to a groubpy object
I also recommend reading about Python's lambda functions:
https://docs.python.org/3/reference/expressions.html

Iterate through two dataframes and create a dictionary one data frame that is a substring in strings found in the second dataframe (values)

I have two dataframes. One is very large and has over 4 million rows of data while the other has about 26k. I'm trying to create a dictionary where the keys are the strings of the smaller data frame. This dataframe (df1) contains substrings or incomplete names and the larger dataframe (df2) contains full names/strings and I want to check if if the substring from df1 is in strings in df2 and then create my dict.
No matter what I try, my code takes long and I keep looking for faster ways to iterate through the df's.
org_dict={}
for rowi in df1.itertuples():
part = rowi.part_name
full_list = []
for rowj in df2.itertuples():
if part in rowj.full_name:
full_list.append(full_name)
org_dict[part]=full_list
Am I missing a break or is there a faster way to iterate through really large dataframes of way over 1 million rows?
Sample data:
df1
part_name
0 aaa
1 bb
2 856
3 cool
4 man
5 a0
df2
full_name
0 aaa35688d
1 coolbbd
2 8564578
3 coolaaa
4 man4857684
5 a03567
expected output:
{'aaa':['aaa35688d','coolaaa'],
'bb':['coolbbd'],
'856':['8564578']
...}
etc

The issue here is that nested for loops perform very badly time-wise as the data grows larger. Luckily, pandas allows us to perform vectorised operations across rows/columns.
I can't properly test without having access to a sample of your data, but I believe this does the trick and performs much faster:
org_dict = {substr: df2.full_name[df2.full_name.str.contains(substr)].tolist() for substr in df1.part_name}

Assign value to dataframe from another dataframe based on two conditions

I am trying to assign values from a column in df2['values'] to a column df1['values']. However values should only be assigned if:
df2['category'] is equal to the df1['category'] (rows are part of the same category)
df1['date'] is in df2['date_range'] (date is in a certain range for a specific category)
So far I have this code, which works, but is far from efficient, since it takes me two days to process the two dfs (df1 has ca. 700k rows).
for i in df1.category.unique():
for j in df2.category.unique():
if i == j: # matching categories
for ia, ra in df1.loc[df1['category'] == i].iterrows():
for ib, rb in df2.loc[df2['category'] == j].iterrows():
if df1['date'][ia] in df2['date_range'][ib]:
df1.loc[ia, 'values'] = rb['values']
break
I read that I should try to avoid using for-loops when working with dataframes. List comprehensions are great, however since I do not have a lot of experience yet, I struggle formulating more complicated code.
How can I iterate over this problem more efficient? What essential key aspect should I think about when iterating over dataframes with conditions?
The code above tends to skip some rows or assigns them wrongly, so I need to do a cleanup afterwards. And the biggest problem, that it is really slow.
Thank you.
Some df1 insight:
df1.head()
date category
0 2015-01-07 f2
1 2015-01-26 f2
2 2015-01-26 f2
3 2015-04-08 f2
4 2015-04-10 f2
Some df2 insight:
df2.date_range[0]
DatetimeIndex(['2011-11-02', '2011-11-03', '2011-11-04', '2011-11-05',
'2011-11-06', '2011-11-07', '2011-11-08', '2011-11-09',
'2011-11-10', '2011-11-11', '2011-11-12', '2011-11-13',
'2011-11-14', '2011-11-15', '2011-11-16', '2011-11-17',
'2011-11-18'],
dtype='datetime64[ns]', freq='D')
df2 other two columns:
df2[['values','category']].head()
values category
0 01 f1
1 02 f1
2 2.1 f1
3 2.2 f1
4 03 f1

Edit: Corrected erroneous code and added OP input from a comment
Alright so if you want to join the dataframes on similar categories, you can merge them :
import pandas as pd
df3 = df1.merge(df2, on = "category")
Next, since date is a timestamp and the "date_range" is actually generated from two columns, per OP's comment, we rather use :
mask = (df3["startdate"] <= df3["date"]) & (df3["date"] <= df3["enddate"])
subset = df3.loc[mask]
Now we get back to df1 and merge on the common dates while keeping all the values from df1. This will create NaN for the subset values where they didn't match with df1 in the earlier merge.
As such, we set df1["values"] where the entries in common are not NaN and we leave them be otherwise.
common_dates = df1.merge(subset, on = "date", how= "left") # keeping df1 values
df1["values"] = np.where(common_dates["values_y"].notna(),
common_dates["values_y"], df1["values"])
N.B : If more than one df1["date"] matches with the date range, you'll have to drop some values otherwise duplicates mess up the explanation.

You could accomplish the first point:
1. df2['category'] is equal to the df1['category']
with the use of a join.
You could then use a for loop for filtering the data poings from df1[date] inside the merged dataframe that are not contemplated in the df2[date_range]. Unfortunately I need more information about the content of df1[date] and df2[date_range] to write the code here that would exactly do that.

How to add multiple new columns to existing csv file without mentioning the column name in python?

I want to add multiple columns in an existing csv. My data looks like this:
50451 51151 53266
100 100 100
1 1 1
where the data starting with (50...) are the columns and below them are the rows.
I have another dataset which looks similar to this:
50014 50013 54567
50 100 100
I am using this code to change it into csv:
df.to_csv('fort.csv', index = False)
but what it does is , it replaces the old columns with new ones. Since , I have to add multiple columns , I can't use df['50014'] everytime. If you guys could suggest something , I would greatly appreciate it.

You can use merge suffixes, to achieve this. In case there are columns with the same names the suffixes will fix the problem.
suffixes : tuple of (str, str), default (‘_x’, ‘_y’)
Suffix to apply to overlapping column names in the left and right side, respectively. To raise an exception on overlapping columns use (False, False).
r = pd.merge(df, df, right_index=True, left_index=True, suffixes=('_left', '_right'), how='outer')
print(r)
For simplicity, I took the same df again:
50451 51151 53266
0 100 100 100
1 1 1 1
50451_left 51151_left 53266_left 50451_right 51151_right 53266_right
0 100 100 100 100 100 100
1 1 1 1 1 1 1
Important is to use the outer join.
Merge is same as join except join may be handy because it is by default used on indices (indexes). They share the same code base ;).

How do I groupby a dataframe based on values that are common to multiple columns?

I am trying to aggregate a dataframe based on values that are found in two columns. I am trying to aggregate the dataframe such that the rows that have some value X in either column A or column B are aggregated together.
More concretely, I am trying to do something like this. Let's say I have a dataframe gameStats:
awayTeam homeTeam awayGoals homeGoals
Chelsea Barca 1 2
R. Madrid Barca 2 5
Barca Valencia 2 2
Barca Sevilla 1 0
... and so on
I want to construct a dataframe such that among my rows I would have something like:
team goalsFor goalsAgainst
Barca 10 5
One obvious solution, since the set of unique elements is small, is something like this:
for team in teamList:
aggregateDf = gameStats[(gameStats['homeTeam'] == team) | (gameStats['awayTeam'] == team)]
# do other manipulations of the data then append it to a final dataframe
However, going through a loop seems less elegant. And since I have had this problem before with many unique identifiers, I was wondering if there was a way to do this without using a loop as that seems very inefficient to me.

The solution is 2 folds, first compute goals for each team when they are home and away, then combine them. Something like:
goals_when_away = gameStats.groupby(['awayTeam'])['awayGoals', 'homeGoals'].agg('sum').reset_index().sort_values('awayTeam')
goals_when_home = gameStats.groupby(['homeTeam'])['homeGoals', 'awayGoals'].agg('sum').reset_index().sort_values('homeTeam')
then combine them
np_result = goals_when_away.iloc[:, 1:].values + goals_when_home.iloc[:, 1:].values
pd_result = pd.DataFrame(np_result, columns=['goal_for', 'goal_against'])
result = pd.concat([goals_when_away.iloc[:, :1], pd_result], axis=1, ignore_index=True)
Note .values when summing to get result in numpy array, and ignore_index=True when concat, these are to avoid pandas trap when it sums by column and index names.