Pandas Replace Values with Dictionary

Question

I have a data frame with the below structure:

        Ranges  Relative_17-Aug  Relative_17-Sep  Relative_17-Oct
0   (0.0, 0.1]  1372             1583             1214
1   (0.1, 0.2]  440              337              648
2   (0.2, 0.3]  111              51               105
3   (0.3, 0.4]  33               10               19
4   (0.4, 0.5]  16               4                9
5   (0.5, 0.6]  7                7                1
6   (0.6, 0.7]  4                3                0
7   (0.7, 0.8]  5                1                0
8   (0.8, 0.9]  2                3                0
9   (0.9, 1.0]  2                0                1
10  (1.0, 2.0]  6                0                2

I am trying to replace column ranges with a dictionary using the below code but it is not working, any hints if I am doing something wrong:

mydict= {"(0.0, 0.1]":"<=10%","(0.1, 0.2]":">10% and <20%","(0.2, 0.3]":">20% and <30%", "(0.3, 0.4]":">30% and <40%", "(0.4, 0.5]":">40% and <50%", "(0.5, 0.6]":">50% and <60%", "(0.6, 0.7]":">60% and <70%", "(0.7, 0.8]":">70% and <80%", "(0.8, 0.9]":">80% and <90%", "(0.9, 1.0]":">90% and <100%", "(1.0, 2.0]":">100%"}
t_df["Ranges"].replace(mydict,inplace=True)

Thanks!

Perhaps casting might help t_df["Ranges"].astype(str).replace(mydict,inplace=True) — Bharath M Shetty
– Bharath M Shetty, Commented Mar 24, 2018 at 15:10
What about using map method to use the value from the dict.? — Sreeram TP
– Sreeram TP, Commented Mar 24, 2018 at 15:15
I got ranges by using groupby ranges = [0,.10,0.20,0.30,0.40,0.50,0.60,0.70,0.80,0.90,1.0,2.0] df_relative.groupby(pd.cut(df_relative["Relative_"+str(month)], ranges)).count() — Rehan Azher
– Rehan Azher, Commented Mar 24, 2018 at 15:17
I have put my full code on pastebin: pastebin.com/embed_iframe/2MqRZ68r — Rehan Azher
– Rehan Azher, Commented Mar 24, 2018 at 15:22

jezrael · Accepted Answer · 2018-03-24 15:13:46Z

2

I think here is best use parameter labels in time of create Ranges column in cut:

labels = ['<=10%','>10% and <20%', ...]
#change by your bins
bins = [0,0.1,0.2...]
t_df['Ranges'] = pd.cut(t_df['col'], bins=bins, labels=labels)

If not possible, cast to string should help as suggest @Dark in comments, for better performance use map:

t_df["Ranges"] = t_df["Ranges"].astype(str).map(mydict)

answered Mar 24, 2018 at 15:13

jezrael

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4 Comments

Rehan Azher Over a year ago

Jezrael above tip works but I am still wondering why replace is not working. I used labels in pd.cut and output is as I required. Thanks

jezrael Over a year ago

@RehanAzher - there is problem it is new IntervalIndex, so possible solution is cast to strings. But cut + labels is better solution.

Rehan Azher Over a year ago

Yes, cut + label is better solution.

Rehan Azher Over a year ago

Delima how to accept both as solution , both are working :)

Sreeram TP · Accepted Answer · 2018-03-24 15:32:04Z

2

By using map function this can be achieved easily and in a straight forward manner as shown below..

mydict= {"(0.0, 0.1]":"<=10%","(0.1, 0.2]":">10% and <20%","(0.2, 0.3]":">20% and <30%", "(0.3, 0.4]":">30% and <40%", "(0.4, 0.5]":">40% and <50%", "(0.5, 0.6]":">50% and <60%", "(0.6, 0.7]":">60% and <70%", "(0.7, 0.8]":">70% and <80%", "(0.8, 0.9]":">80% and <90%", "(0.9, 1.0]":">90% and <100%", "(1.0, 2.0]":">100%"}

t_df["Ranges"] = t_df["Ranges"].map(lambda x : mydict[str(x)])

Hope this helps..!!

edited Mar 24, 2018 at 15:32

answered Mar 24, 2018 at 15:23

Sreeram TP

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7 Comments

BENY Over a year ago

t_df["Ranges"].map( mydict)

Sreeram TP Over a year ago

Yep, I was just using lambda's it is more like an habit..!!

Rehan Azher Over a year ago

Below worked : ` t_df["Ranges"] = t_df["Ranges"].map(lambda x : mydict[str(x)]) `

Sreeram TP Over a year ago

Good catch, I was using str ranges for trying out the code.

Rehan Azher Over a year ago

Sorry, I had to accept @Jezrael answer as the solution not only he/she provided a solution but right away knew how i got the ranges.

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