Python/Pandas Subtract Only if Value is not 0 - python-3.x

Question

I'm starting with data that looks something like this, but with a lot more rows:
Location Sample a b c d e f g h i
1 w 14.6 0 0 0 0 0 0 0 16.8
2 x 0 13.6 0 0 0 0 0 0 16.5
3 y 0 0 15.5 0 0 0 0 0 16.9
4 z 0 0 0 0 14.3 0 0 0 15.7
...
The data is indexed by the first two columns. I need to subtract the values in column i from each of the values in a - h, adding a new column to the right of the data frame for each original column. However, if there is a zero in the first column, I want it to stay zero instead of subtracting. For example, if my code worked I would have the following columns added to the data frame on the right
Location Sample ... a2 b2 c2 d2 e2 f2 g2 h2
1 w ... -2.2 0 0 0 0 0 0 0
2 x ... 0 -2.9 0 0 0 0 0 0
3 y ... 0 0 -1.4 0 0 0 0 0
4 z ... 0 0 0 0 -1.4 0 0 0
...
I'm trying to use where in pandas to only subtract the value in column i if the value in the current column is not zero using the following code:
import pandas as pd
normalizer = i
columns = list(df.columns.values)
for column in columns:
if column == normalizer: continue
newcol = gene + "2"
df[newcol] = df.where(df[column] == 0,
df[column] - df[normalizer], axis = 0)
I'm using a for loop because the number of columns will not always be the same, and the column that is being subtracted will have a different name using different data sets.
I'm getting this error: "ValueError: Wrong number of items passed 9, placement implies 1".
I think the subtraction is causing the issue, but I can't figure out how to change it to make it work. Any assistance would be greatly appreciated.
Thanks in advance.

Answer 1

Method 1 (pretty fast: roughly 3 times faster than method 2)
1. Select columns that is relavent
2. Do subtraction
3. Elementwise mutiplication with a 0, 1 matrix that constructed before the substraction. Each element in (df_ref > 0) is 0 if it was originally 0 and 1 otherwise.
ith_col = df["i"]
subdf = df.iloc[:, 2:-1] # a - h columns
df_temp = subdf.sub(ith_col, axis=0).multiply(subdf > 0).add(0)
df_temp.columns = ['a2', 'b2', 'c2', 'd2', 'e2', 'f2', 'g2', 'h2'] # rename columns
df_desired = pd.concat([df, df_temp], axis=1)
Note in this method, the 0 is negative. Thus, we have an extra add(0) in the end. Yes, a 0 can be negative. :P
Method 2 (more readable)
1. Find the greater than 0 part with a condition.
2. Select rows that is relavent
3. Substract
4. Fill in 0.
ith_col = df["i"]
df[df > 0].iloc[:,2:-1].sub(ith_col, axis=0).fillna(0)
The second method is pretty similar to #Wen's answer. Credits to him :P
Speed comparison of two methods (tested on Python 3 and pandas 0.20)
%timeit subdf.sub(ith_col, axis=0).multiply(subdf > 0).add(0)
688 µs ± 30.9 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit df[df > 0].iloc[:,2:-1].sub(ith_col, axis=0).fillna(0)
2.97 ms ± 248 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Reference:
DataFrame.multiply perform elementwise multiplication with another data frame.

Answer 2

Using mask + fillna
df.iloc[:,2:-1]=df.iloc[:,2:-1].mask(df.iloc[:,2:-1]==0).sub(df['i'],0).fillna(0)
df
Out[116]:
Location Sample a b c d e f g h i
0 1 w -2.2 0.0 0.0 0.0 0.0 0.0 0.0 0.0 16.8
1 2 x 0.0 -2.9 0.0 0.0 0.0 0.0 0.0 0.0 16.5
2 3 y 0.0 0.0 -1.4 0.0 0.0 0.0 0.0 0.0 16.9
3 4 z 0.0 0.0 0.0 0.0 -1.4 0.0 0.0 0.0 15.7
Update
normalizer = ['i','Location','Sample']
df.loc[:,~df.columns.isin(normalizer)]=df.loc[:,~df.columns.isin(normalizer)].mask(df.loc[:,~df.columns.isin(normalizer)]==0).sub(df['i'],0).fillna(0)