Subtract one column from another in pandas - with a condition - python

I have this code that will subtract, for each person (AAC or AAB), timepoint 1 from time point 2 data.
i.e this is the original data:
pep_seq AAC-T01 AAC-T02 AAB-T01 AAB-T02
0 0 1 2.0 NaN 4.0
1 4 3 2.0 6.0 NaN
2 4 3 NaN 6.0 NaN
3 4 5 2.0 6.0 NaN
This is the code:
import sys
import numpy as np
from sklearn.metrics import auc
import pandas as pd
from numpy import trapz
#read in file
df = pd.DataFrame([[0,1,2,np.nan,4],[4,3,2,6,np.nan],[4,3,np.nan,6,np.nan],[4,5,2,6,np.nan]],columns=['pep_seq','AAC-T01','AAC-T02','AAB-T01','AAB-T02'])
#standardise the data by taking T0 away from each sample
df2 = df.drop(['pep_seq'],axis=1)
df2 = df2.apply(lambda x: x.sub(df2[x.name[:4]+"T01"]))
df2.insert(0,'pep_seq',df['pep_seq'])
print(df)
print(df2)
This is the output (i.e. df2)
pep_seq AAC-T01 AAC-T02 AAB-T01 AAB-T02
0 0 0 1.0 NaN NaN
1 4 0 -1.0 0.0 NaN
2 4 0 NaN 0.0 NaN
3 4 0 -3.0 0.0 NaN
...but what I actually wanted was to subtract the T01 columns from all the others EXCEPT for when the T01 value is NaN in which case keep the original value, so the desired output was (see the 4.0 in AAB-T02):
pep_seq AAC-T01 AAC-T02 AAB-T01 AAB-T02
0 0 0 1.0 NaN 4.0
1 4 0 -1.0 0 NaN
2 4 0 NaN 0 NaN
3 4 0 -3.0 0 NaN
Could someone show me where I went wrong? Note that in real life, there are ~100 timepoints per person, not just two.

You can fill the nan to 0 when doing subtraction
df2 = df2.apply(lambda x: x.sub(df2[x.name[:4]+"T01"].fillna(0)))
# ^^^^ Changes here
df2.insert(0,'pep_seq',df['pep_seq'])
print(df2)
pep_seq AAC-T01 AAC-T02 AAB-T01 AAB-T02
0 0 0 1.0 NaN 4.0
1 4 0 -1.0 0.0 NaN
2 4 0 NaN 0.0 NaN
3 4 0 -3.0 0.0 NaN

I hope that I understand you correctly but numpy.where() should do it for you.
Have a look here: condition based substraction

Related

How would you flip and fold diagonaly a matrix with pandas?

I have some datas I would like to organize for visualization and statistics but I don't know how to proceed.
The data are in 3 columns (stimA, stimB and subjectAnswer) and 10 rows (numero of pairs) and they are from a pairwise comparison test, in panda's dataFrame format. Example :
stimA
stimB
subjectAnswer
1
2
36
3
1
55
5
3
98
...
...
...
My goal is to organize them as a matrix with each row and column corresponding to one stimulus with the subjectAnswer data grouped to the left side of the matrix' diagonal (in my example, the subjectAnswer 36 corresponding to stimA 1 and stimB 2 should go to the index [2][1]), like this :
stimA/stimB
1
2
3
4
5
1
...
2
36
3
55
4
...
5
...
...
98
I succeeded in pivoting the first table to the matrix but I couldn't succeed the arrangement on the left side of the diag of my datas, here is my code :
import pandas as pd
import matplotlib.pyplot as plt
import numpy as np
session1 = pd.read_csv(filepath, names=['stimA', 'stimB', 'subjectAnswer'])
pivoted = session1.pivot('stimA','stimB','subjectAnswer')
Which gives :
session1 :
stimA stimB subjectAnswer
0 1 3 6
1 4 3 21
2 4 5 26
3 2 3 10
4 1 2 6
5 1 5 6
6 4 1 6
7 5 2 13
8 3 5 15
9 2 4 26
pivoted :
stimB 1 2 3 4 5
stimA
1 NaN 6.0 6.0 NaN 6.0
2 NaN NaN 10.0 26.0 NaN
3 NaN NaN NaN NaN 15.0
4 6.0 NaN 21.0 NaN 26.0
5 NaN 13.0 NaN NaN NaN
The expected output for pivoted :
stimB 1 2 3 4 5
stimA
1 NaN NaN Nan NaN NaN
2 6.0 NaN Nan NaN NaN
3 6.0 10.0 NaN NaN NaN
4 6.0 26.0 21.0 NaN NaN
5 6.0 13.0 15.0 26.0 NaN
Thanks a lot for your help !
If I understand you correctly, the stimuli A and B are interchangeable. So to get the matrix layout you want, you can swap A with B in those rows where A is smaller than B. In other words, you don't use the original A and B for the pivot table, but the maximum and minimum of A and B:
session1['stim_min'] = np.min(session1[['stimA', 'stimB']], axis=1)
session1['stim_max'] = np.max(session1[['stimA', 'stimB']], axis=1)
pivoted = session1.pivot('stim_max', 'stim_min', 'subjectAnswer')
pivoted
stim_min 1 2 3 4
stim_max
2 6.0 NaN NaN NaN
3 6.0 10.0 NaN NaN
4 6.0 26.0 21.0 NaN
5 6.0 13.0 15.0 26.0
sort the columns stimA and stimB along the columns axis and assign two temporary columns namely x and y in the dataframe. Here sorting is required because we need to ensure that the resulting matrix clipped on the upper right side.
Pivot the dataframe with index as y, columns as x and values as subjectanswer, then reindex the reshaped frame in order to ensure that all the available unique stim names are present in the index and columns of the matrix
session1[['x', 'y']] = np.sort(session1[['stimA', 'stimB']], axis=1)
i = np.union1d(session1['x'], session1['y'])
session1.pivot('y', 'x','subjectAnswer').reindex(i, i)
x 1 2 3 4 5
y
1 NaN NaN NaN NaN NaN
2 6.0 NaN NaN NaN NaN
3 6.0 10.0 NaN NaN NaN
4 6.0 26.0 21.0 NaN NaN
5 6.0 13.0 15.0 26.0 NaN

How to interpolate in Pandas using only previous values?

This is my dataframe:
df = pd.DataFrame(np.array([ [1,5],[1,6],[1,np.nan],[2,np.nan],[2,8],[2,4],[2,np.nan],[2,10],[3,np.nan]]),columns=['id','value'])
id value
0 1 5
1 1 6
2 1 NaN
3 2 NaN
4 2 8
5 2 4
6 2 NaN
7 2 10
8 3 NaN
This is my expected output:
id value
0 1 5
1 1 6
2 1 7
3 2 NaN
4 2 8
5 2 4
6 2 2
7 2 10
8 3 NaN
This is my current output using this code:
df.value.interpolate(method="krogh")
0 5.000000
1 6.000000
2 9.071429
3 10.171429
4 8.000000
5 4.000000
6 2.357143
7 10.000000
8 36.600000
Basically, I want to do two important things here:
Groupby ID then Interpolate using only above values not below row values
This should do the trick:
df["value_interp"]=df.value.combine_first(df.groupby("id")["value"].apply(lambda y: y.expanding().apply(lambda x: x.interpolate(method="krogh").to_numpy()[-1], raw=False)))
Outputs:
id value value_interp
0 1.0 5.0 5.0
1 1.0 6.0 6.0
2 1.0 NaN 7.0
3 2.0 NaN NaN
4 2.0 8.0 8.0
5 2.0 4.0 4.0
6 2.0 NaN 0.0
7 2.0 10.0 10.0
8 3.0 NaN NaN
(It interpolates based only on the previous values within the group - hence index 6 will return 0 not 2)
You can group by id and then loop over groups to make interpolations. For id = 2 interpolation will not give you value 2
import pandas as pd
import numpy as np
df = pd.DataFrame(np.array([ [1,5],[1,6],[1,np.nan],[2,np.nan],[2,8],[2,4],[2,np.nan],[2,10],[3,np.nan]]),columns=['id','value'])
data = []
for name, group in df.groupby('id'):
group_interpolation = group.interpolate(method='krogh', limit_direction='forward', axis=0)
data.append(group_interpolation)
df = (pd.concat(data)).round(1)
Output:
id value
0 1.0 5.0
1 1.0 6.0
2 1.0 7.0
3 2.0 NaN
4 2.0 8.0
5 2.0 4.0
6 2.0 4.7
7 2.0 10.0
8 3.0 NaN
Current pandas.Series.interpolate does not support what you want so to achieve your goal you need to do 2 grouby's that will account for your desire to use only previous rows. The idea is as follows: to combine into one group only missing value (!!!) and previous rows (it might have limitations if you have several missing values in a row, but it serves well for your toy example)
Suppose we have a df:
print(df)
ID Value
0 1 5.0
1 1 6.0
2 1 NaN
3 2 NaN
4 2 8.0
5 2 4.0
6 2 NaN
7 2 10.0
8 3 NaN
Then we will combine any missing values within a group with previous rows:
df["extrapolate"] = df.groupby("ID")["Value"].apply(lambda grp: grp.isnull().cumsum().shift().bfill())
print(df)
ID Value extrapolate
0 1 5.0 0.0
1 1 6.0 0.0
2 1 NaN 0.0
3 2 NaN 1.0
4 2 8.0 1.0
5 2 4.0 1.0
6 2 NaN 1.0
7 2 10.0 2.0
8 3 NaN NaN
You may see, that when grouped by ["ID","extrapolate"] the missing value will fall into the same group as nonnull values of previous rows.
Now we are ready to do extrapolation (with spline of order=1):
df.groupby(["ID","extrapolate"], as_index=False).apply(lambda grp:grp.interpolate(method="spline",order=1)).drop("extrapolate", axis=1)
ID Value
0 1.0 5.0
1 1.0 6.0
2 1.0 7.0
3 2.0 NaN
4 2.0 8.0
5 2.0 4.0
6 2.0 0.0
7 2.0 10.0
8 NaN NaN
Hope this helps.

Forward fill missing values by group after condition is met in pandas

I'm having a bit of trouble with this. My dataframe looks like this:
id amount dummy
1 130 0
1 120 0
1 110 1
1 nan nan
1 nan nan
2 nan 0
2 50 0
2 20 1
2 nan nan
2 nan nan
So, what I need to do is, after the dummy gets value = 1, I need to fill the amount variable with zeroes for each id, like this:
id amount dummy
1 130 0
1 120 0
1 110 1
1 0 nan
1 0 nan
2 nan 0
2 50 0
2 20 1
2 0 nan
2 0 nan
I'm guessing I'll need some combination of groupby('id'), fillna(method='ffill'), maybe a .loc or a shift() , but everything I tried has had some problem or is very slow. Any suggestions?
The way I will use
s = df.groupby('id')['dummy'].ffill().eq(1)
df.loc[s&df.dummy.isna(),'amount']=0
You can do this much easier:
data[data['dummy'].isna()]['amount'] = 0
This will select all the rows where dummy is nan and fill the amount column with 0.
IIUC, ffill() and mask the still-nan:
s = df.groupby('id')['amount'].ffill().notnull()
df.loc[df['amount'].isna() & s, 'amount'] = 0
Output:
id amount dummy
0 1 130.0 0.0
1 1 120.0 0.0
2 1 110.0 1.0
3 1 0.0 NaN
4 1 0.0 NaN
5 2 NaN 0.0
6 2 50.0 0.0
7 2 20.0 1.0
8 2 0.0 NaN
9 2 0.0 NaN
Could you please try following.
df.loc[df['dummy'].isnull(),'amount']=0
df
Output will be as follows.
id amount dummy
0 1 130.0 0.0
1 1 120.0 0.0
2 1 110.0 1.0
3 1 0.0 NaN
4 1 0.0 NaN
5 2 NaN 0.0
6 2 50.0 0.0
7 2 20.0 1.0
8 2 0.0 NaN
9 2 0.0 NaN

Concatenating crosstabs of different variables

I have a Pandas (0.23.4) DataFrame with several categorical columns.
df = pd.DataFrame(np.random.choice([True, False, np.nan], (6,4)), columns = ['a','b','c','d'])
a b c d
0 NaN 1.0 NaN NaN
1 NaN 1.0 NaN 0.0
2 1.0 NaN 1.0 NaN
3 0.0 NaN 0.0 1.0
4 NaN 1.0 NaN NaN
5 NaN 1.0 0.0 1.0
I have two sets of columns of interest:
cross_cols = ['a', 'b']
type_cols = ['c', 'd']
I would like to get a cross tab of counts of each cross_col variable with each type_col variable (a with c and d, and b with c and d), excluding NaN, all displayed side-by-side. The desired result is:
c d
0.0 1.0 All 0.0 1.0 All
a 0.0 0 0 0 1 1 2
1.0 2 1 3 1 0 1
All 2 1 3 2 1 3
b 0.0 0 0 0 0 1 1
1.0 2 1 3 2 0 2
All 2 1 3 2 1 3
Notice that I am not interested in counts for different combinations of a and b or of c and d, which is what I'm getting by changing the index and columns parameters of pd.crosstab.
Currently I'm using the following code:
cross_rows = []
for col in cross_cols:
cross_rows.append(pd.concat([pd.crosstab(df[col], df[type_var],margins=True) for type_var in type_cols],axis=1,keys = type_cols,sort=True))
results = pd.concat(cross_rows, keys = cross_cols,sort=True)
It gives the following result:
c d
c 0.0 1.0 All 0.0 1.0 All
a 1.0 2.0 1.0 3.0 1 0 1
All 2.0 1.0 3.0 2 1 3
0.0 NaN NaN NaN 1 1 2
b 1.0 2.0 1.0 3.0 2 0 2
All 2.0 1.0 3.0 2 1 3
0.0 NaN NaN NaN 0 1 1
The result is fine, but the code is slow and a bit ugly. I suspect that there's a faster and more Pythonic approach. Is there a single function call that would get the job done, or another faster solution?

Shift NaNs to the end of their respective rows

I have a DataFrame like :
0 1 2
0 0.0 1.0 2.0
1 NaN 1.0 2.0
2 NaN NaN 2.0
What I want to get is
Out[116]:
0 1 2
0 0.0 1.0 2.0
1 1.0 2.0 NaN
2 2.0 NaN NaN
This is my approach as of now.
df.apply(lambda x : (x[x.notnull()].values.tolist()+x[x.isnull()].values.tolist()),1)
Out[117]:
0 1 2
0 0.0 1.0 2.0
1 1.0 2.0 NaN
2 2.0 NaN NaN
Is there any efficient way to achieve this ? apply Here is way to slow .
Thank you for your assistant!:)
My real data size
df.shape
Out[117]: (54812040, 1522)
Here's a NumPy solution using justify -
In [455]: df
Out[455]:
0 1 2
0 0.0 1.0 2.0
1 NaN 1.0 2.0
2 NaN NaN 2.0
In [456]: pd.DataFrame(justify(df.values, invalid_val=np.nan, axis=1, side='left'))
Out[456]:
0 1 2
0 0.0 1.0 2.0
1 1.0 2.0 NaN
2 2.0 NaN NaN
If you want to save memory, assign it back instead -
df[:] = justify(df.values, invalid_val=np.nan, axis=1, side='left')
Your best easiest option is to use sorted on df.apply/df.transform and sort by nullity.
df = df.apply(lambda x: sorted(x, key=pd.isnull), 1)
df
0 1 2
0 0.0 1.0 2.0
1 1.0 2.0 NaN
2 2.0 NaN NaN
You may also pass np.isnan to the key argument.

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