'How to check if float pandas column contains only integer numbers?
I have a dataframe
df = pd.DataFrame(data=np.arange(10),columns=['v']).astype(float)
How to make sure that the numbers in v are whole numbers?
I am very concerned about rounding/truncation/floating point representation errors
Solution 1:[1]
Here's a simpler, and probably faster, approach:
(df[col] % 1 == 0).all()
To ignore nulls:
(df[col].fillna(-9999) % 1 == 0).all()
Solution 2:[2]
If you want to check multiple float columns in your dataframe, you can do the following:
col_should_be_int = df.select_dtypes(include=['float']).applymap(float.is_integer).all()
float_to_int_cols = col_should_be_int[col_should_be_int].index
df.loc[:, float_to_int_cols] = df.loc[:, float_to_int_cols].astype(int)
Keep in mind that a float column, containing all integers will not get selected if it has np.NaN values. To cast float columns with missing values to integer, you need to fill/remove missing values, for example, with median imputation:
float_cols = df.select_dtypes(include=['float'])
float_cols = float_cols.fillna(float_cols.median().round()) # median imputation
col_should_be_int = float_cols.applymap(float.is_integer).all()
float_to_int_cols = col_should_be_int[col_should_be_int].index
df.loc[:, float_to_int_cols] = float_cols[float_to_int_cols].astype(int)
Solution 3:[3]
For completeness, Pandas v1.0+ offer the convert_dtypes() utility, that (among 3 other conversions) performs the requested operation for all dataframe-columns (or series) containing only integer numbers.
If you wanted to limit the conversion to a single column only, you could do the following:
>>> df.dtypes # inspect previous dtypes
v float64
>>> df["v"] = df["v"].convert_dtype()
>>> df.dtypes # inspect converted dtypes
v Int64
Solution 4:[4]
On 27 331 625 rows it works well. Time : 1.3sec
df['is_float'] = df[field_fact_qty]!=df[field_fact_qty].astype(int)
This way took Time : 4.9s
df[field_fact_qty].apply(lambda x : (x.is_integer()))
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | scott |
| Solution 2 | |
| Solution 3 | ankostis |
| Solution 4 |
