Question

我的熊猫数据框如下，

    Location    Loc_Code    Service_Onroad  Service_Cargo
0   Bangalore   Vadakara01  Common  Common
1   Bangalore   Hosdung02   Transit Transit
2   Bangalore   Hosdung02   Common  Transit
3   Bangalore   Kozhikode   Common  Common
4   Bangalore   Malappuram  Transit Transit
5   Bangalore   Kalpetta09  Common  Common
6   Bangalore   Kasargod    Transit Transit
7   Bangalore   Maananthavaadi  Common  Common
8   Bangalore   Alattur Transit Transit
9   Bangalore   Pazhani Common  NaN
10  Bangalore   Ernakulam   Transit Transit
11  Bangalore   Aluva   Connect Connect
12  Bangalore   Kochi   Common  Transit
13  Bangalore   Kochi   Transit Transit
14  Bangalore   Paravoor    Common  Common
15  Bangalore   Chaavakkaad Transit Transit
16  Bangalore   Tirur   Transit Transit
17  Bangalore   Mannaarkkad Common  Common
18  Bangalore   Koramangala NaN NaN
19  Bangalore   Rajna   NaN NaN
20  Bangalore   Palai09 Interconnect    NaN
21  Bangalore   Palai09 Common  NaN

我想将Service_Onroad与Service_Cargo进行比较，如果匹配，我想在新列中将其写为“ Matched”（即结果）。如果不匹配，我想将其写为“ Not Matched”。在任何列（Service_Onroad和Service_Cargo中，如果值是NaN，那么我不想在那写任何东西。

我们如何进行比较？我相信在 Pandas 中，Nan也被用作比较的字符串，从而导致“ FALSE”？那么我们如何排除呢？

Answer 1

尝试一下：

df['Match'] = np.where((df['Service_Onroad']==df['Service_Cargo']), 'Matched', 'Non-Matched')
df['Match'] = np.where((df['Service_Onroad'].isna()) | (df['Service_Cargo'].isna()), '', df['Match'])
print(df)

     Location        Loc_Code Service_Onroad Service_Cargo        Match
0   Bangalore      Vadakara01         Common        Common      Matched
1   Bangalore       Hosdung02        Transit       Transit      Matched
2   Bangalore       Hosdung02         Common       Transit  Non-Matched
3   Bangalore       Kozhikode         Common        Common      Matched
4   Bangalore      Malappuram        Transit       Transit      Matched
5   Bangalore      Kalpetta09         Common        Common      Matched
6   Bangalore        Kasargod        Transit       Transit      Matched
7   Bangalore  Maananthavaadi         Common        Common      Matched
8   Bangalore         Alattur        Transit       Transit      Matched
9   Bangalore         Pazhani         Common           NaN
10  Bangalore       Ernakulam        Transit       Transit      Matched
11  Bangalore           Aluva        Connect       Connect      Matched
12  Bangalore           Kochi         Common       Transit  Non-Matched
13  Bangalore           Kochi        Transit       Transit      Matched
14  Bangalore        Paravoor         Common        Common      Matched
15  Bangalore     Chaavakkaad        Transit       Transit      Matched
16  Bangalore           Tirur        Transit       Transit      Matched
17  Bangalore     Mannaarkkad         Common        Common      Matched
18  Bangalore     Koramangala            NaN           NaN
19  Bangalore           Rajna            NaN           NaN
20  Bangalore         Palai09   Interconnect           NaN
21  Bangalore         Palai09         Common           NaN

Answer 2

在结果列中定义要满足的条件：

condition_1=((df['Service_Onroad'].isnull())|(df['Service_Cargo'].isnull()))
condition_2=df['Service_Onroad']==df['Service_Cargo']

然后使用“ np.where”并嵌套条件以获得结果：

df['result']=np.where(condition_1,'',(np.where(condition_2,'match','no_match')))

df.to_excel('PATH',na_rep='')

进行比较时如何忽略Nan

2 个答案: