英文:
How do you sort a dataframe with the integer in a column with strings and integers on every row?
问题
df['a'] = df['a'].str.extract('(\d+)').astype(int)df = df.sort_values(by=['a'], ignore_index=True)
英文:
How would you sort the following dataframe:
df = pd.DataFrame({'a':['abc_1.2.6','abc_1.2.60','abc_1.2.7','abc_1.2.9','abc_1.3.0','abc_1.3.10','abc_1.3.100','abc_1.3.11'], 'b':[1,2,3,4,5,6,7,8]})>>>a b0 abc_1.2.6 11 abc_1.2.60 22 abc_1.2.7 33 abc_1.2.9 44 abc_1.3.0 55 abc_1.3.10 66 abc_1.3.100 77 abc_1.3.11 8
to achieve this output?
>>>a b0 abc_1.2.6 11 abc_1.2.7 32 abc_1.2.9 43 abc_1.2.60 24 abc_1.3.0 55 abc_1.3.10 66 abc_1.3.11 87 abc_1.3.100 7
I understand that integers in strings can be accessed through string transformations, however I'm unsure how to handle this in a dataframe. Obviously df.sort_values(by=['a'],ignore_index=True) is unhelpful in this case.
答案1
得分: 2
One way to use is natsorted with iloc :
#pip install natsortfrom natsort import natsortedout = df.iloc[natsorted(range(len(df)), key=lambda x: df.loc[x, "a"])]
Or even shorter, as suggested by @Stef, use natsort_key as a key of sort_values :
from natsort import natsort_keyout = df.sort_values(by="a", key=natsort_key, ignore_index=True)
Output :
print(out)a b0 abc_1.2.6 11 abc_1.2.7 32 abc_1.2.9 43 abc_1.2.60 24 abc_1.3.0 55 abc_1.3.10 66 abc_1.3.11 87 abc_1.3.100 7
英文:
One way to use is natsorted with iloc :
#pip install natsortfrom natsort import natsortedout = df.iloc[natsorted(range(len(df)), key=lambda x: df.loc[x, "a"])]
Or even shorter, as suggested by @Stef, use natsort_key as a key of sort_values :
from natsort import natsort_keyout = df.sort_values(by="a", key=natsort_key, ignore_index=True)
Output :
print(out)a b0 abc_1.2.6 11 abc_1.2.7 32 abc_1.2.9 43 abc_1.2.60 24 abc_1.3.0 55 abc_1.3.10 66 abc_1.3.11 87 abc_1.3.100 7
答案2
得分: 1
你可以在排序之前对值应用key函数：
df = (df.sort_values(by=['a'], ignore_index=True,key=lambda x: x.map(lambda v:tuple(map(int, v[4:].split('.'))))))
a b0 abc_1.2.6 11 abc_1.2.7 32 abc_1.2.9 43 abc_1.2.60 24 abc_1.3.0 55 abc_1.3.10 66 abc_1.3.11 87 abc_1.3.100 7
英文:
You can apply the key function to the values before sorting:
df = (df.sort_values(by=['a'], ignore_index=True,key=lambda x: x.map(lambda v:tuple(map(int, v[4:].split('.'))))))
a b0 abc_1.2.6 11 abc_1.2.7 32 abc_1.2.9 43 abc_1.2.60 24 abc_1.3.0 55 abc_1.3.10 66 abc_1.3.11 87 abc_1.3.100 7
答案3
得分: 1
这是另一种方法，使用str.findall()和explode()
df.sort_values('a', key=lambda x: x.str.findall(r'\d+').explode().astype(int).groupby(level=0).agg(tuple))
输出：
a b0 abc_1.2.6 12 abc_1.2.7 33 abc_1.2.9 41 abc_1.2.60 24 abc_1.3.0 55 abc_1.3.10 67 abc_1.3.11 86 abc_1.3.100 7
英文:
Here is another way by using str.findall() and explode()
df.sort_values('a',key = lambda x: x.str.findall(r'\d+').explode().astype(int).groupby(level=0).agg(tuple))
Output:
a b0 abc_1.2.6 12 abc_1.2.7 33 abc_1.2.9 41 abc_1.2.60 24 abc_1.3.0 55 abc_1.3.10 67 abc_1.3.11 86 abc_1.3.100 7
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