Check if value in column is an integer and return True-CodePudding

I have a dataframe with a quantity column. How do I check which values in that column are integers?

I have tried

if df['quantity'].map(type) == int:
        True
else:
        item_no = df['item_number'].tolist()
        print(item_no)

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

Ideally, I would like the output to be a list of the item numbers where the quantity is not an integer.

CodePudding user response：

If you want to check if there are any Python ints in your column, try this:

df['quantity'].map(type).eq(int).any()

CodePudding user response：

Assuming you just want the indices where the value is an int, the following should suffice df.index[(df['quantity'].map(type)==int)].tolist()

CodePudding user response：

If you want to identify the non integers you can use:

df[df['quantity'].map(type).ne(int)]

df[~df['quantity'].apply(lambda x: isinstance(int, x))]

Or for non numbers:

df[pd.to_numeric(df['quantity'], errors='coerce').isna()]

CodePudding user response：

The following code returns a list of the item numbers for where the 'quantity' column does not contain an integer

q = df.loc[~df['quantity'].str.isdigit(), 'item_num'].tolist()
print(q)