I have a dataframe that looks like
df
viz a1_count a1_mean a1_std
n 3 2 0.816497
y 0 NaN NaN
n 2 51 50.000000
I want to convert the "viz" column to 0 and 1, based on a conditional. I've tried:
df['viz'] = 0 if df['viz'] == "n" else 1
but I get:
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
You're trying to compare a scalar with the entire series which raise the ValueError you saw. A simple method would be to cast the boolean series to int:
In [84]:
df['viz'] = (df['viz'] !='n').astype(int)
df
Out[84]:
viz a1_count a1_mean a1_std
0 0 3 2 0.816497
1 1 0 NaN NaN
2 0 2 51 50.000000
You can also use np.where:
In [86]:
df['viz'] = np.where(df['viz'] == 'n', 0, 1)
df
Out[86]:
viz a1_count a1_mean a1_std
0 0 3 2 0.816497
1 1 0 NaN NaN
2 0 2 51 50.000000
Output from the boolean comparison:
In [89]:
df['viz'] !='n'
Out[89]:
0 False
1 True
2 False
Name: viz, dtype: bool
And then casting to int:
In [90]:
(df['viz'] !='n').astype(int)
Out[90]:
0 0
1 1
2 0
Name: viz, dtype: int32
pd.to_numeric(myDF['myDFCell'], errors='coerce'). This is probably newer pandas syntax. The coerce flag tells it to convert that which cannot be converted to number to NA so it won't throw errors.
NaN so this is slightly different use case hereFrom @TMWP's comment above:
pd.to_numeric(myDF['myDFCell'], errors='coerce')
It works like a charm and is a quick and simple one liner
