If I create a dataframe like so:
import pandas as pd, numpy as np
df = pd.DataFrame(np.random.randint(0,100,size=(100, 2)), columns=list('AB'))
How would I change the entry in column A to be the number 16 from row 0 -15, for example? In other words, how do I replace cells based purely on index?
Use loc:
df.loc[0:15,'A'] = 16
print (df)
A B
0 16 45
1 16 5
2 16 97
3 16 58
4 16 26
5 16 87
6 16 51
7 16 17
8 16 39
9 16 73
10 16 94
11 16 69
12 16 57
13 16 24
14 16 43
15 16 77
16 41 0
17 3 21
18 0 98
19 45 39
20 66 62
21 8 53
22 69 47
23 48 53
Solution with ix is deprecated.
In addition to the other answers, here is what you can do if you have a list of individual indices:
indices = [0,1,3,6,10,15]
df.loc[indices,'A'] = 16
print(df.head(16))
Output:
A B
0 16 4
1 16 4
2 4 3
3 16 4
4 1 1
5 3 0
6 16 4
7 2 1
8 4 4
9 3 4
10 16 0
11 3 1
12 4 2
13 2 2
14 2 1
15 16 1
Very interesting observation, that code below does change the value in the original dataframe
df.loc[0:15,'A'] = 16
But if you use a pretty similar code like this
df.loc[0:15]['A'] = 16
Than it will give back just a copy of your dataframe with changed value and doesn't change the value in the original df object. Hope that this will save some time for someone dealing with this issue.
One more solution is
df.at[0:15, 'A']=16
print(df.head(20))
OUTPUT:
A B
0 16 44
1 16 86
2 16 97
3 16 79
4 16 94
5 16 24
6 16 88
7 16 43
8 16 64
9 16 39
10 16 84
11 16 42
12 16 8
13 16 72
14 16 23
15 16 28
16 18 11
17 76 15
18 12 38
19 91 6
Could you instead of 16, update the value of that column to -1.0? for me, it returns 255 instead of -1.0.
>>> effect_df.loc[3:5, ['city_SF', 'city_Seattle']] = -1.0
Rent city_SF city_Seattle
0 3999 1 0
1 4000 1 0
2 4001 1 0
3 3499 255 255
4 3500 255 255
5 3501 255 255
6 2499 0 1
7 2500 0 1
8 2501 0 1
To Mad Physicist: it appears that at first you need to change the column data types from short integer to float. Looks like your -1.0 was cast as short integer.
