I have a pandas dataframe defined as:
A B SUM_C
1 1 10
1 2 20
I would like to do a cumulative sum of SUM_C and add it as a new column to the same dataframe. In other words, my end goal is to have a dataframe that looks like below:
A B SUM_C CUMSUM_C
1 1 10 10
1 2 20 30
Using cumsum in pandas on group() shows the possibility of generating a new dataframe where column name SUM_C is replaced with cumulative sum. However, my ask is to add the cumulative sum as a new column to the existing dataframe.
Thank you
Just apply cumsum on the pandas.Series df['SUM_C'] and assign it to a new column:
df['CUMSUM_C'] = df['SUM_C'].cumsum()
Result:
df
Out[34]:
A B SUM_C CUMSUM_C
0 1 1 10 10
1 1 2 20 30
Related
I am trying to do some aggregation of fields and I get an additional row on the aggregate column- in this case 'sum'. How do I get rid of this? This makes it difficult to process this dataframe in subsequent processing
import pandas as pd
Grade=[['A',1],['A',2],['A',10],['B',4],['B',2],['C',3],['D',10],['D',5],['D',1]]
Grade_df=pd.DataFrame(Grade,columns=['grade','count'])
Grade_Aggregate=Grade_df.groupby(['grade']).agg({'count':['sum']}).reset_index()
Grade_Aggregate
grade count
sum
0 A 13
1 B 6
2 C 3
3 D 16
You can use Groupby.sum here directly.
Grade_df.groupby('grade', as_index=False)['count'].sum()
grade count
0 A 13
1 B 6
2 C 3
3 D 16
If you want to remove the 'Sum':
Grade_Aggregate.columns=[i[0] for i in Grade_Aggregate.columns]
If you want it to be appended to the column name:
Grade_Aggregate.columns=["_".join(i) for i in Grade_Aggregate.columns]
I can't figure out how (DataFrame - Groupby) works.
Specifically, given the following dataframe:
df = pd.DataFrame([['usera',1,100],['usera',5,130],['userc',1,100],['userd',5,100]])
df.columns = ['id','date','sum']
id date sum
0 usera 1 100
1 usera 5 130
2 userc 1 100
3 userd 5 100
Passing the below code returns:
df['shift'] = df['date']-df.groupby(['id'])['date'].shift(1)
id date sum shift
0 usera 1 100
1 usera 5 130 4.0
2 userc 1 100
3 userd 5 100
How did Python know that I meant for it to match by id column?
It doesn't even appear in df['date']
Let us dissect the command df['shift'] = df['date']-df.groupby(['id'])['date'].shift(1).
df['shift'] appends a new column "shift" in the dataframe.
df['date'] returns Series using date column from the dataframe.
0 1
1 5
2 1
3 5
Name: date, dtype: int64
df.groupby(['id'])['date'].shift(1) groupby(['id']) creates a groupby object.
From that groupby object selecting date column and shifting one (previous) value using shift(1). By the way, this also a Series.
df.groupby(['id'])['date'].shift(1)
0 NaN
1 1.0
2 NaN
3 NaN
Name: date, dtype: float64
The Series obtained from step 3 is subtracted (element-wise) with the Series obtained from Step 2. The result is assigned to the df['shift'] column.
df['date']-df.groupby(['id'])['date'].shift(1)
0 NaN
1 4.0
2 NaN
3 NaN
Name: date, dtype: float64
I am not exactly knowing what you are trying, but groupby() method is usuful if you have several same objects in a column (like you usera) and you want to calculate for example the sum(), mean(), find max() etc. of all columns or just one specific column.
e.g. df.groupby(['id'])['sum'].sum() groups you usera and just select the sum column and build the sum over all usera. So it is 230. If you would use .mean() it would output 115 etc. And it also does it for all other unique id in your id column. In the example from above it outputs one column with just three rows (user a-c).
Greetz, miGa
What I wanna do:
Column 'angle' has tracked about 20 angles per second (can vary). But my 'Time' timestamp has only an accuracy of 1s (therefore always about ~20 rows are having the same timestamp)(total rows of over 1 million in the dataframe).
My result shall be a new dataframe with a changing timestamp for each row. The angle for the timestamp shall be the median of the ~20 timestamps in that intervall.
My Idea:
I iterate through the rows and check if the timestamp has changed.
If so, I select all timestamps until it changes, calculate the median, and append it to a new dataframe.
Nevertheless I have many many big data files and I am wondering if there is a faster way to achieve my goal.
Right now my code is the following (see below).
It is not fast and I think there must be a better way to do that with pandas/numpy (or something else?).
a = 0
for i in range(1,len(df1.index)):
if df1.iloc[[a],[1]].iloc[0][0]==df1.iloc[[i],[1]].iloc[0][0]:
continue
else:
if a == 0:
df_result = df1[a:i-1].median()
else:
df_result = df_result.append(df1[a:i-1].median(), ignore_index = True)
a = i
You can use groupby here. Below, I made a simple dummy dataframe.
import pandas as pd
df1 = pd.DataFrame({'time': [1,1,1,1,1,1,2,2,2,2,2,2],
'angle' : [8,9,7,1,4,5,11,4,3,8,7,6]})
df1
time angle
0 1 8
1 1 9
2 1 7
3 1 1
4 1 4
5 1 5
6 2 11
7 2 4
8 2 3
9 2 8
10 2 7
11 2 6
Then, we group by the timestamp and take the median of the angle column within that group, and convert the result to a pandas dataframe.
df2 = pd.DataFrame(df1.groupby('time')['angle'].median())
df2 = df2.reset_index()
df2
time angle
0 1 6.0
1 2 6.5
You can use the .agg after grouping function to select operation according to the column
df1.groupby('Time', as_index=False).agg({"angle":"median"})
This is very similar to this question, except I want my code to be able to apply to the length of a dataframe, instead of specific columns.
I have a DataFrame, and I'm trying to get a sum of each row to append to the dataframe as a column.
df = pd.DataFrame([[1,0,0],[20,7,1],[63,13,5]],columns=['drinking','drugs','both'],index = ['First','Second','Third'])
drinking drugs both
First 1 0 0
Second 20 7 1
Third 63 13 5
Desired output:
drinking drugs both total
First 1 0 0 1
Second 20 7 1 28
Third 63 13 5 81
Current code:
df['total'] = df.apply(lambda row: (row['drinking'] + row['drugs'] + row['both']),axis=1)
This works great. But what if I have another dataframe, with seven columns, which are not called 'drinking', 'drugs', or 'both'? Is it possible to adjust this function so that it applies to the length of the dataframe? That way I can use the function for any dataframe at all, with a varying number of columns, not just a dataframe with columns called 'drinking', 'drugs', and 'both'?
Something like:
df['total'] = df.apply(for col in df: [code to calculate sum of each row]),axis=1)
You can use sum:
df['total'] = df.sum(axis=1)
If you need sum only some columns, use subset:
df['total'] = df[['drinking', 'drugs', 'both']].sum(axis=1)
what about something like this :
df.loc[:, 'Total'] = df.sum(axis=1)
with the output :
Out[4]:
drinking drugs both Total
First 1 0 0 1
Second 20 7 1 28
Third 63 13 5 81
It will sum all columns by row.
I am trying to do a division of column 0 by columns 1 and 2. From the below, I would like to return a dataframe of 10 rows, 3 columns. The first column should all be 1's. Instead I get a 10x10 dataframe. What am I doing wrong?
data = np.random.randn(10,3)
df = pd.DataFrame(data)
df[0] / df
First you should create a 10 by 3 DataFrame with all columns equal to the first column and then divide it with your DataFrame.
df[[0, 0, 0]] / df.values
or
df[[0, 0, 0]].values / df
If you want to keep the column names.
(I use .values to avoid reindexing which will fail due to duplicate column values.)
You need to match the dimension of the Series with the rows of the DataFrame. There are a few ways to do this but I like to use transposes.
data = np.random.randn(10,3)
df = pd.DataFrame(data)
(df[0] / df.T).T
0 1 2
0 1 -0.568096 -0.248052
1 1 -0.792876 -3.539075
2 1 -25.452247 1.434969
3 1 -0.685193 -0.540092
4 1 0.451879 -0.217639
5 1 -2.691260 -3.208036
6 1 0.351231 -1.467990
7 1 0.249589 -0.714330
8 1 0.033477 -0.004391
9 1 -0.958395 -1.530424