我的目标很简单。
从下面的数据框（df3）：
如果mr_no等于定义的数字（我有下面的列表）然后开火，如果fire1等于覆盖然后开火，如果fire2等于主要或次要然后开火，否则如果这三个都没有符合标准而不是火。
DF3：
mr_no fire1 fire2
0 A5155 covered primary
1 FS2328 covered neither
3 133158 not_covered primary
4 FS2328 not_covered neither
5 A5128 not_covered neither
6 A5128 not_covered neither
7 133158 covered primary
8 133158 not_covered neither
9 A5128 not_covered neither
10 A5129 not_covered neither
我尝试了以下内容来提取我需要的代码：
mr = ["A5115","133158","FS2114","FS2110","FS2115","FS2111","FS2149","FS2150","FS2017","FS2018"]
def fire(row):
if row['fire1'] == ('covered') + row['fire2'] == ('primary') | ('secondary') + row['mr_no'] == (mr):
return 'fire'
return 'not fire'
df3['fire_final'] = df3.apply (lambda row: fire(row),axis=1)
但是，我的结果没有找到，但我知道应该有一些东西：
IN: df3.groupby('fire_final').ssno.count()
OUT: fire_final
not fire 214166
Name: ssno, dtype: int64
我认为我的错误与我的列表有关，但我的编程知识仍然是新手所以我不确定如何为列表做条件。
与往常一样，我非常感谢您的帮助。
更新：
尝试edChum推荐的代码会产生结果，但正如你所看到的那样，行120044和120045显示fire1被“覆盖”而fire2是主要的，但是它们不是火，但它们应该显示为火。
df3['fire_final'] = np.where((df3['mr_no'].isin(mr)) & (df3['fire1'] == 'covered') & (df3['fire2'].isin(['primary','secondary'])), 'fire', 'not fire')
mr_no fire1 fire2 fire_final
120020 203670 not_covered neither not fire
120021 001500 not_covered neither not fire
120022 5273 not_covered neither not fire
120023 190365 not_covered neither not fire
120024 A5155 not_covered neither not fire
120025 236707 not_covered neither not fire
120026 A9088 not_covered primary not fire
120027 A9088 not_covered primary not fire
120028 FS1476 not_covered primary not fire
120029 A9092 not_covered primary not fire
120031 FS1476 not_covered neither not fire
120032 FS1476 not_covered neither not fire
120033 AFM002 not_covered primary not fire
120034 A9092 not_covered primary not fire
120035 FS1476 not_covered primary not fire
120036 FS1476 not_covered primary not fire
120037 007881 not_covered neither not fire
120038 007286 not_covered neither not fire
120039 A010 not_covered neither not fire
120041 N5012 not_covered neither not fire
120042 FS1478 covered primary fire
120043 FS1478 covered primary fire
120044 A2075 covered primary not fire
120045 136492 covered primary not fire
更新：
当我使用|时，EdChum的推荐工作而不是＆amp;
df3['fire_final'] = np.where((df3['mr_no'].isin(mr)) | (df3['fire1'] == 'covered') | (df3['fire2'].isin(['primary','secondary'])), 'fire', 'not fire')
产生预期的结果：
mr_no fire1 fire2 fire_final
120062 FS0937 not_covered secondary fire
120063 FS0937 not_covered secondary fire
120064 FS0900 not_covered secondary fire
120066 FS2037 covered primary fire
120067 A9090 not_covered neither not fire
120068 129440 covered neither fire
120069 FS1629 covered primary fire
120070 FS2037 covered primary fire
120071 FS2037 covered primary fire
120072 A9074 not_covered neither not fire
120073 ABE102 not_covered neither not fire
120074 ABE100 not_covered neither not fire
120075 A5125 not_covered neither not fire
120076 199424 not_covered neither not fire
120077 179873 not_covered neither not fire
120078 A021 not_covered neither not fire
120079 A022 not_covered neither not fire
120080 A022 not_covered neither not fire
120081 A022 not_covered neither not fire
答案 0 :(得分：2)
Python不是C ++也不是C.因此，它使用了一些不同的结构。
在您的情况下，当您希望在if为真时或 thing1为真时触发thing2语句，您使用or不是|。
在Python中|执行按位or操作。
所以你想要这样的东西：
if (something) or (something):
return 'fired'
答案 1 :(得分：1)
我相信你可以使用np.where：
In [6]:
mr = ["A5115","133158","FS2114","FS2110","FS2115","FS2111","FS2149","FS2150","FS2017","FS2018"]
df['fire_final'] = np.where((df['mr_no'].isin(mr)) & (df['fire1'] == 'covered') & (df['fire2'].isin(['primary','secondary'])), 'fire', 'not fire')
df
Out[6]:
mr_no fire1 fire2 fire_final
0 A5155 covered primary not fire
1 FS2328 covered neither not fire
3 133158 not_covered primary not fire
4 FS2328 not_covered neither not fire
5 A5128 not_covered neither not fire
6 A5128 not_covered neither not fire
7 133158 covered primary fire
8 133158 not_covered neither not fire
9 A5128 not_covered neither not fire
10 A5129 not_covered neither not fire
目前还不清楚你真正想要的是什么，但如果任何条件为真，你想要返回true，那么你需要或条件：
df['fire_final'] = np.where((df['mr_no'].isin(mr)) | (df['fire1'] == 'covered') | (df['fire2'].isin(['primary','secondary'])), 'fire', 'not fire')
需要注意的事项是，对于数组，您需要对&，|和{{1}使用按位运算符~，and，or这是因为我们在这里比较一个数组与一些布尔条件，我们想要一个返回的布尔数组。对于多个条件，由于运算符优先级，您需要将条件括在not中。