Replacing values greater than a number in pandas dataframe

Question

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

Answer 1

Very simply : df[df > 9] = 11

Answer 2

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]

Answer 3

I came for a solution to replacing each element larger than h by 1 else 0, which has the simple solution:

df = (df > h) * 1

(This does not solve the OP's question as all df <= h are replaced by 0.)

Answer 4

You can use numpy indexing, accessed through the .values function.

df['col'].values[df['col'].values > x] = y

where you are replacing any value greater than x with the value of y.

So for the example in the question:

df1['A'].values[df1['A'] > 9] = 11