Trying to divide a dataframe column by a float yields NaN

Question

Background I deal with a csv datasheet that prints out columns of numbers. I am working on a program that will take the first column, ask a user for a time in float (ie. 45 and a half hours = 45.5) and then subtract that number from the first column. I have been successful in that regard. Now, I need to find the row index of the "zero" time point. I use min to find that index and then call that off of the following column A1. I need to find the reading at Time 0 to then normalize A1 to so that on a graph, at the 0 time point the reading is 1 in column A1 (and eventually all subsequent columns but baby steps for me)

time_zero = float(input("Which time would you like to be set to 0?"))
df['A1']= df['A1']-time_zero

This works fine so far to set the zero time.

zero_location_series = df[df['A1'] == df['A1'].min()]
r1 = zero_location_series[' A1.1']
df[' A1.1'] = df[' A1.1']/r1

Here's where I run into trouble. The first line will correctly identify a series that I can pull off of for all my other columns. Next r1 correctly identifies the proper A1.1 value and this value is a float when I use type(r1). However when I divide df[' A1.1']/r1 it yields only one correct value and that value is where r1/r1 = 1. All other values come out NaN.

My Questions:

1. How to divide a column by a float I guess? Why am I getting NaN?
2. Is there a faster way to do this as I need to do this for 16 columns.(ie 'A2/r2' 'a3/r3' etc.)
3. Do I need to do inplace = True anywhere to make the operations stick prior to resaving the data? or is that only for adding/deleting rows?

Example

Dataframe that looks like this !https://i.stack.imgur.com/fQxTM.png zero time sets properly (image not shown)

after dividing the column !https://i.stack.imgur.com/dA3te.png

Answer 1

If you want to divide every value in the column by r1 it's best to apply, for example:

import pandas as pd
df = pd.DataFrame([1,2,3,4,5])
# apply an anonymous function to the first column ([0]), divide every value
# in the column by 3
df = df[0].apply(lambda x: x/3.0, 0)
print(df)

So you'd probably want something like this:

df = df["A1.1"].apply(lambda x: x/r1, 0)

This really only answers part 2 of you question. Apply is probably your best bet for running a function on multiple rows and columns quickly. As for why you're getting nans when dividing by a float, is it possible the values in your columns are anything other than floats or integers?

Answer 2

This should work:

df['A1.1']=df['A1.1']/df['A1.1'].min()

I think the reason df[' A1.1'] = df[' A1.1']/r1 did not work was because r1 is a series. Try r1? instead of type(r1) and pandas will tell you that r1 is a series, not an individual float number.

To do it in one attempt, you have to iterate over each column, like this:

for c in df:
    df[c] = df[c]/df[c].min()