ValueError: Invalid format specifier when using f-strings inside a function

Question

I am trying to format a pandas series: for all values, I want to add thousands separators and limit decimal places to two. In addition, for negative values, instead of a - sign, I want to enclose it with parentheses. So output should look like '20,000.00', '(152,111.12)'etc.

I know the f-string method works because when I run

val = 20000
f'{val:,.2f}'

It gives me the correct result '20,000.00'. But when I tried to create a function and apply it to an entire column in Pandas:

def format_pnl(value):
    # Format the value as a string with thousand separators and 2 decimal places
    formatted_value = f"{value:, .2f}"
    # If the value is negative, wrap it in parentheses
    if value < 0:
        formatted_value = f"({formatted_value})"
    return formatted_value

I got the ValueError: Invalid format specifier. This is really mind boggling to me since the format specifier absolutely works. What am I missing here?

Are you trying to apply this to the dataframe directly or after extracting data from it? — SimonT
– SimonT, Commented Dec 23, 2022 at 20:13

Scott Hunter · Accepted Answer · 2022-12-23 20:08:56Z

1

You added a space between the comma and period.

answered Dec 23, 2022 at 20:08

Scott Hunter

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1 Comment

Bowen Liu Over a year ago

My goodness. So that's it. Thanks Scott, It solved the problem. I was under the wrong impression that spaces don't matter in the formatting. Is space forbidden in python string formatting? I thought since I have ':,' and '.2f' then it should work. Thanks again.

SimonT · Accepted Answer · 2022-12-23 20:38:14Z

1

You can do this quite easily and apply it to your specific column

import pandas as pd

df = pd.DataFrame([100000, 1113, 510101.05, 6456464, -846464.12])

def format_pnl(value):
    return f"{value:,.2f}" if value >= 0 else f"({abs(value):,.2f})"

# Specify the specific column you want here
df[0] = df[0].apply(format_pnl)

display(df)

    0
0   100,000.00
1   1,113.00
2   510,101.05
3   6,456,464.00
4   (846,464.12)

edited Dec 23, 2022 at 20:38

answered Dec 23, 2022 at 20:06

SimonT

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1 Comment

Scott Hunter Over a year ago

I don't see pd defined in the posted code.

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