I have a large dataframe which looks as:
df1['A'].ix[1:3] 2017-01-01 02:00:00 [33, 34, 39] 2017-01-01 03:00:00 [3, 43, 9]
I want to replace each element greater than 9 with 11.
So, the desired output for above example is:
df1['A'].ix[1:3] 2017-01-01 02:00:00 [11, 11, 11] 2017-01-01 03:00:00 [3, 11, 9]
Edit:
My actual dataframe has about 20,000 rows and each row has list of size 2000.
Is there a way to use numpy.minimum function for each row? I assume that it will be faster than list comprehension method?
Advertisement
Answer
You can use apply with list comprehension:
df1['A'] = df1['A'].apply(lambda x: [y if y <= 9 else 11 for y in x])
print (df1)
A
2017-01-01 02:00:00 [11, 11, 11]
2017-01-01 03:00:00 [3, 11, 9]
Faster solution is first convert to numpy array and then use numpy.where:
a = np.array(df1['A'].values.tolist())
print (a)
[[33 34 39]
[ 3 43 9]]
df1['A'] = np.where(a > 9, 11, a).tolist()
print (df1)
A
2017-01-01 02:00:00 [11, 11, 11]
2017-01-01 03:00:00 [3, 11, 9]