我想根據(jù)索引中的值創(chuàng)建一個列：如果索引以字母值開頭而不是 'I0'，則返回“P”，否則返回“C”。嘗試過：df['new_col'] = ['P' if (x[0].isalpha() and not x[0].startswith("I0")) else 'C' for x in df.index]但它返回以以下開頭的行的“P” 'I0'： A B C new_colIndex I00001 1.325337 4.692308 1.615385 PI00002 1.614780 3.615385 0.769231 PI00003 1.141453 5.461538 2.000000 PI00004 0.918300 8.538462 2.769231 PI00005 1.189606 11.846154 2.692308 PI00006 0.941459 7.153846 2.153846 PI00007 0.466383 12.153846 9.384615 PI00008 0.308627 198.692308 23.461538 PI00011 0.537142 23.384615 6.846154 PI00012 1.217390 11.923077 1.230769 PI00013 1.052840 3.384615 2.000000 P...可重現(xiàn)的例子：df = pd.DataFrame({'A': {'I00001': 1.3253365856660808, 'I00002': 1.6147800817881086, 'I00003': 1.1414534979918203, 'I00004': 0.9183004454646491, 'I00005': 1.1896061362142527, 'I00006': 0.941459102789141, 'I00007': 0.46638312473267185, 'I00008': 0.3086270976042302, 'I00011': 0.5371419441302684, 'I00012': 1.2173904641254587, 'I00013': 1.052839529263679, 'I00014': 1.3587324409735149, 'I00015': 3.464101615137755, 'I00016': 1.1989578808281798, 'I00018': 0.2433560755649686, 'I00019': 0.5510000980337852, 'I00020': 3.464101615137755, 'I00022': 1.0454523047666737, 'I00023': 1.3850513878332371, 'I00024': 1.3314720972390754}, 'B': {'I00001': 4.6923076923076925, 'I00002': 3.6153846153846154, 'I00003': 5.461538461538462, 'I00004': 8.538461538461538, 'I00005': 11.846153846153847, 'I00006': 7.153846153846154, 'I00007': 12.153846153846153, 'I00008': 198.69230769230768, 'I00011': 23.384615384615383, 'I00012': 11.923076923076923, 'I00013': 3.3846153846153846, 'I00014': 1.0,)
2 回答
30秒到達戰(zhàn)場
TA貢獻1828條經(jīng)驗 獲得超6個贊
非循環(huán)解決方案numpy.where：
df['new_col'] = np.where(df.index.str[0].str.isalpha() &
? ? ? ? ? ? ? ? ? ? ? ? ?~df.index.str.startswith("I0"), 'P', 'C')
您的解決方案 -x[0]從中刪除x[0].startswith("I0")- 它測試第一個值，如果不是I0，則始終是True：
df['new_col'] = ['P' if (x[0].isalpha() and not x.startswith("I0"))? ?
? ? ? ? ? ? ? ? ? ? ?else 'C' for x in df.index]
測試：
df = pd.DataFrame({'A': {'AA00001': 1.3253365856660808,
? 'I00002': 1.6147800817881086,
? 'IR0003': 1.1414534979918203,
? '00004': 0.9183004454646491,
? '**00005': 1.1896061362142527,
? 'I00007': 0.46638312473267185}}
)
df['new_col'] = np.where(df.index.str[0].str.isalpha() &
? ? ? ? ? ? ? ? ? ? ? ? ?~df.index.str.startswith("I0"), 'P', 'C')
df['new_col1'] = ['P' if (x[0].isalpha() and not x.startswith("I0"))? ?
? ? ? ? ? ? ? ? ? ? ? else 'C' for x in df.index]
print (df)
? ? ? ? ? ? ? ? A new_col new_col1
**00005? 1.189606? ? ? ?C? ? ? ? C
00004? ? 0.918300? ? ? ?C? ? ? ? C
AA00001? 1.325337? ? ? ?P? ? ? ? P
I00002? ?1.614780? ? ? ?C? ? ? ? C
I00007? ?0.466383? ? ? ?C? ? ? ? C
IR0003? ?1.141453? ? ? ?P? ? ? ? P
翻閱古今
TA貢獻1780條經(jīng)驗 獲得超5個贊
您正在檢查x[0].startswith("I0")您的代碼，這是不正確的嘗試這個（檢查x.startswith("I0")）
df['new_col'] = ['P' if (x[0].isalpha() and not x.startswith("I0")) else 'C' for x in df.index]添加回答
舉報
0/150
提交
取消