CODEWITHSUNDEEP
pythondatabasepandas
Zanam
Zanam

Reputation: 4807

Replacing values greater than a number in pandas dataframe

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?

Upvotes: 63

Views: 175806

Answers (5)

kpetrou
kpetrou

Reputation: 441

I know this is an old post, but pandas now supports DataFrame.where directly. In your example:

df.where(df <= 9, 11, inplace=True)

Please note that pandas' where is different than numpy.where. In pandas, when the condition == True, the current value in the dataframe is used. When condition == False, the other value is taken.

EDIT:

You can achieve the same for just a column with Series.where:

df['A'].where(df['A'] <= 9, 11, inplace=True)

Upvotes: 41

CFW
CFW

Reputation: 304

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.)

Upvotes: 6

D.Griffiths
D.Griffiths

Reputation: 2317

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

Upvotes: 26

Edouard Cuny
Edouard Cuny

Reputation: 1260

Very simply : df[df > 9] = 11

Upvotes: 65

jezrael
jezrael

Reputation: 862731

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]

Upvotes: 47

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