How to truncate two columns of decimal points in a python pandas dataframe?-CodePudding

I need help with truncating columns of decimals within a dataset, that ties in with other calcuations to output a 1 or 0 binary column called "scenario1".

I'm basically two columns (columns A and columns B) of decimals along a time-based index. The columns may have varied amount of decimal points.

e.g. ColA can be 4 decimals, ColB can be 6 decimals.

ColA	ColB
0.9954	0.995642
0.9854	0.997450

If the value of ColA and ColB is close enough, I want to output TRUE. To do that, I have to somehow truncate both ColA and ColB to one decimal place without any rounding up or down. So it becomes the following:

ColA	ColB
0.9	0.9
0.9	0.9

I need this truncation to happen within a function "scenario1", trying to have the code be efficient as possible. My current failed attempts are those lines with math.trunc(1000 * df.tenkan)/1000.

CodePudding user response：

You can use numpy trunc:

df = pd.DataFrame({'ColA ': [0.9954, 0.9854], 'ColB': [0.995642, 0.99745]})

np.trunc(10 * df) / 10

Result:

   ColA   ColB
0    0.9   0.9
1    0.9   0.9

CodePudding user response：

I think the easiest way to do this would be to use applymap paired with the truncate function of the math module. Here is an example:

trunc = lambda x: math.trunc(10 * x)/10

df.applymap(trunc)

You'll need to apply this over your columns of interest, but i tested it on a few arbitrary examples and it worked well. Hope that helps! Can expound on detail if necessary.

CodePudding user response：

You could also make use of regular expression:

df = pd.DataFrame({'ColA ': [0.99989, 0.986767], 'ColB': [0.9890, 0.9588]})

func = lambda x: re.match(r'\d .\d{1}', str(x)).group(0)
df.applymap(func)

Alternatively here is the not so elegant approach where you first convert the number into a string, then you get the different parts of the string separately and lastly covert the string back to a float(Yes this is not efficient) :

def func(x): 
  # Convert number to a string 
  digits = str(x).split(".")

  # Manually put the number back together: 
  digit = digits[0]   "."   digits[1][:1]
  return float(digit)


df.applymap(func)

Results:

ColA  | ColB
0.9   |  0.9
0.9   |  0.9