2 回答

TA贡献1807条经验 获得超9个赞
使用嵌套列表理解:
df = pd.DataFrame({'rnti':['0x00000034,0x0000008a,0x00000145','0x00000034,0x0000008a']})
print (df)
rnti
0 0x00000034,0x0000008a,0x00000145
1 0x00000034,0x0000008a
rnti_list = [[int(idx, 16) for idx in x.split(",")] for x in df['rnti']]
print (rnti_list)
[[52, 138, 325], [52, 138]]
df['new'] = [[int(idx, 16) for idx in x.split(",")] for x in df['rnti']]
print (df)
rnti new
0 0x00000034,0x0000008a,0x00000145 [52, 138, 325]
1 0x00000034,0x0000008a [52, 138]

TA贡献1852条经验 获得超7个赞
使用 apply
前任:
import pandas as pd
df = pd.DataFrame({"rnti": ["0x00000034,0x0000008a,0x00000145"]})
print( df["rnti"].str.split(",").apply(lambda x: [int(i, 16) for i in x]) )
输出:
0 [52, 138, 325]
Name: rnti, dtype: object
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