如何计算 dataframe 中的尾随零而不计算小数点后的零？

Question

I am trying to calculate the trailing zeros in a dataframe and add them as a new column. However my code:

dataset['Trade price round'] = dataset['Trade Price'].apply(lambda x: len(str(x))) - dataset['Trade Price'].apply(lambda x: len(str(x).rstrip('0')))
print(dataset.tail(10))

is giving me 1 for values that have a zero at the end as well as for numbers that have a trailing decimal point of zero. I only want the prior.

	Bloomberg Code	index...	Trade price	Trade price round
90	ATCOB SS Equity	149...	254.6	0
91	AZN SS Equity	150...	613	1
92	AZN SS Equity	151...	610	1
93	AZN SS Equity	152...	610	1
94	AZN SS Equity	153...	610	1
95	BOL SS Equity	154...	184.5	0
96	BOL SS Equity	155...	182.1	0
97	BOL SS Equity	156...	182.1	0
98	BOL SS Equity	157...	182.1	0
99	ELUXB SS Equity	158...	228	1

As you can see with index number 99 it still counts the floating point whilst index number 94 has exactly what I want. What am I doing wrong?

Doing something like this with another column I have would help:

dataset['Trade price int'] = 1 if dataset['Trade price float'] == 0 else 0

but of course I get the error:

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

How can I apply any() or some other function to this?

Answer 1

Have you tried the int() function so that you can count the trailing zeros from the integer number and not the float one?

Answer 2

I've got the answer. Since the trailing zero stops at the decimal you need to use float then strip both the zero and the decimal point then the zeros again. However this adds two to the number at the end so you need to add an if statement to remove 2 if the difference is greater than zero like this:

dataset['Trade price round'] = dataset['Trade Price'].apply(lambda x: len(str(float(x))) - len(str(float(x)).rstrip('0').rstrip('.').rstrip('0')) - 2 if len(str(float(x))) - len(str(float(x)).rstrip('0').rstrip('.').rstrip('0')) > 0 else 0)