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Dataframe:

 Date and time Transaction Date Transcation currency Card currency Description 0 21.12.2021 14:31 21.12.2021 + 500,00 ₽ + 500,00 ₽ Внутрибанковский перевод с договора 52876***** 1 21.12.2021 15:43 21.12.2021 - 454,00 ₽ - 454,00 ₽ Оплата в KFC AVIAMOTORNAYA.2 Gorod Moskva RUS 2 21.12.2021 22:29 21.12.2021 + 16,98 ₽ + 16,98 ₽ Проценты на остаток 3 21.12.2021 22:56 21.12.2021 + 887,00 ₽ + 887,00 ₽ Кэшбэк за обычные покупки 4 21.12.2021 23:15 21.12.2021 - 59,00 ₽ - 59,00 ₽ Плата за оповещения об операциях .. ... ... ... ... ... 330 19.03.2022 21:32 19.03.2022 + 1 200,00 ₽ + 1 200,00 ₽ Внутрибанковский перевод с договора 5287***** 331 20.03.2022 01:06 21.03.2022 - 50,50 ₽ - 50,50 ₽ Операция в других кредитных организациях QIWI 332 NaN NaN NaN NaN MOSKVA RUS 333 21.03.2022 12:43 21.03.2022 - 10,00 ₽ - 10,00 ₽ Внешний перевод по номеру телефона +79263****** 334 21.03.2022 12:43 21.03.2022 - 30,00 ₽ - 30,00 ₽ Комиссия за внешний перевод

I have a piece of code that iterates over the data frame by column:

for index, row in df.iterrows(): print(row["Date and time"])

I need to delete the rows where it is NaN. How do I do that? I have tried .isnull() and .notnull() but they return Errors

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    df.dropna(inplace=True) Commented Mar 25, 2022 at 14:14

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Use a boolean mask. I assume you loaded the data frame and 'Nan' is read as a string:

df = df[df["Date and time"]!='NaN']
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This won't actually work. You need to do df[df["Date and time"].notna()] because NaN == NaN is always False. Or preferably you could do df = df.dropna(). Also, please don't answer duplicate questions :)

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