如何解决Python中列表理解的多个条件
我想基于索引中的值创建一列：
如果索引以字母值而不是'I0'开头，则返回“ P”，否则返回“ C”。
尝试：
df['new_col'] = ['P' if (x[0].isalpha() and not x[0].startswith("I0")) else 'C' for x in df.index]
但是对于以'I0'开头的行，它返回了'P'：
A B C new_col
Index
I00001 1.325337 4.692308 1.615385 P
I00002 1.614780 3.615385 0.769231 P
I00003 1.141453 5.461538 2.000000 P
I00004 0.918300 8.538462 2.769231 P
I00005 1.189606 11.846154 2.692308 P
I00006 0.941459 7.153846 2.153846 P
I00007 0.466383 12.153846 9.384615 P
I00008 0.308627 198.692308 23.461538 P
I00011 0.537142 23.384615 6.846154 P
I00012 1.217390 11.923077 1.230769 P
I00013 1.052840 3.384615 2.000000 P
...
可复制的示例：
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,'I00015': 0.07692307692307693,'I00016': 0.6153846153846154,'I00018': 481.7692307692308,'I00019': 7.3076923076923075,'I00020': 0.07692307692307693,'I00022': 1.6153846153846154,'I00023': 0.5384615384615384,'I00024': 12.538461538461538},'C': {'I00001': 1.6153846153846154,'I00002': 0.7692307692307693,'I00003': 2.0,'I00004': 2.769230769230769,'I00005': 2.6923076923076925,'I00006': 2.1538461538461537,'I00007': 9.384615384615385,'I00008': 23.46153846153846,'I00011': 6.846153846153846,'I00012': 1.2307692307692308,'I00013': 2.0,'I00014': 0.38461538461538464,'I00016': 0.46153846153846156,'I00018': 79.07692307692308,'I00019': 3.6923076923076925,'I00022': 1.1538461538461537,'I00023': 0.46153846153846156,'I00024': 2.3076923076923075}}
)
解决方法
使用numpy.where的非循环解决方案：
df['new_col'] = np.where(df.index.str[0].str.isalpha() &
~df.index.str.startswith("I0"),'P','C')
您的解决方案-从x[0]中删除x[0].startswith("I0")-如果不是I0，则测试第一个值，始终为True：
df['new_col'] = ['P' if (x[0].isalpha() and not x.startswith("I0"))
else 'C' for x in df.index]
测试：
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"),'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
您正在检查代码中的x_counts= []
for item in df['x']:
item_count = len(df[df['x']==item])
x_counts.append(item_count)
df['x_count'] = x_counts
，这是错误的
改用此方法（检查x[0].startswith("I0")）
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