How can I change every value in a pandas dataframe column?-CodePudding

I have this pandas dataframe including a column for months whose type is int64, and I want to change every value in the column to be the corresponding season of the year.

For example if x is any of 12, 1, or 2, change the value of x to 'winter', etc.

I have tried some for loops but no use. I think there is a one liner that can do this using lambda x.

I supposed that every month has only 1 season so for example winter is in months 12, 1, and 2, and so on.

For example (0, 1, and 2 are indices):

input:

0  1
1  3
2  6

output:

0  winter
1  spring
2  summer

My problem is with month 12 or else I could have used bins.

Any ideas?

CodePudding user response：

One fast and flexible way is to map a dict from current column values to new column values:

d = dict(zip(range(1,13), ["winter"]*2   ["spring"]*3   ["summer"]*3   ["fall"]*3   ["winter"]))    
df = pd.DataFrame(dict(month=[3,4,6,1,5,6,2,4,5]))

df = df.month.map(d)

output:

0    spring
1    spring
2    summer
3    winter
4    spring
5    summer
6    winter
7    spring
8    spring
Name: month, dtype: object

CodePudding user response：

then u can easily write this function

def map_fn(x: int):
    if x in range(3, 6):
        return "spring"
    if x in range(6, 9):
        return "summer"
    if x in range(9, 12):
        return "autumn"
    return "winter"

and then apply it to your column like this

df.seasons.apply(map_fn)

to have this

CodePudding user response：

I solved it using pandas replace and cut methods:

df['month']=df['month'].replace(12,0)

bin_edges = [-1,2,5,8,11]

bin_labels= ["winter",'spring','summer','fall']

df['month']=pd.cut(df['month'], bin_edges, labels=bin_labels)

Thanks all.