条件或Isin方法

Question

我正在自动化一个必要的任务--列person_id，但是我有这两个df。

df_1 = pd.DataFrame({'Code':['100', '101', '102', '103', '104'], 'person_id': ['4000', '4001', '4002', '4003', '4004']})

df_2 = pd.DataFrame ({'Code':['100', '101', '102', '103', '104'],'owner_1':['False', 'True', 'False', 'False','False'], 'owner_2':['True', 'False', 'False', 'False','False']})

我需要的是确认什么是真正的所有者--使用列"Code“创建第三列，并使用条件词创建类似于"Owner_n_id”的列，在其中我收到了person_id列，但我对如何编写该列感到有点困惑，我所做的一切都不是很好。

我需要我的新df如下所示

df_3 = pd.DataFrame ({'Code':['100', '101', '102', '103', '104'], 'owner_1_id':['False', '4001', 'False', 'False', 'False'],'owner_2_id':['4000', 'False', 'False', 'False','False']})

Answer 1

TL;DR

如果这意味着要用于更多的所有者，则可以创建一个循环。

df_3 = df_1.merge(df_2, on = "Code")
NUM_OWNERS = 2
for owner in range(1, NUM_OWNERS+1):
     indexing = df_3[f"owner_{owner}"] == "True"
     df_3.loc[indexing, f"owner_{owner}"] = df_3.loc[indexing, "person_id"]

# finally drop the extra column
df_3.drop("person_id", axis = 1, inplace = True)

我将首先合并1和2，所以第三个更一致：

>>> df_3 = df_1.merge(df_2, on = "Code")
  Code person_id owner_1 owner_2
0  100      4000   False    True
1  101      4001    True   False
2  102      4002   False   False
3  103      4003   False   False
4  104      4004   False   False

一旦我们有了这个，我们就想访问owner_1 == "True"的位置

owner_1_is_true = df_3.owner_1 == "True"
# and know we can update the value based in this indexing:
df_3.owner_1[owner_1_is_true] = df_3.person_id[owner_1_is_true]

# repeat for owner 2:
owner_2_is_true = df_3.owner_2 == "True"
df_3.owner_2[owner_2_is_true] = df_3.person_id[owner_2_is_true]

# drop the `person_id` we created on the merge
df_3.drop("person_id", axis = 1, inplace = True)

结果：

  Code owner_1 owner_2
0  100   False    4000
1  101    4001   False
2  102   False   False
3  103   False   False
4  104   False   False

如果这意味着要用于更多的所有者，则可以创建一个循环。

NUM_OWNERS = 2
for owner in range(1, NUM_OWNERS+1):
     indexing = df_3[f"owner_{owner}"] == "True"
     df_3.loc[indexing, f"owner_{owner}"] = df_3.loc[indexing, "person_id"]

# finally drop the extra column
df_3.drop("person_id", axis = 1, inplace = True)

Answer 2

此代码直接更改df_2。您可以在继续之前将df_2复制到df_3中。请注意，对False和True使用字符串值与使用布尔运算符相比效率低下。

df_2['owner_1_id'] = 'False'
df_2.loc[df_2['owner_1'] == 'True', ['owner_1_id']] = df_2.merge(df_1)['person_id']

df_2['owner_2_id'] = 'False'
df_2.loc[df_2['owner_2'] == 'True', ['owner_2_id']] = df_2.merge(df_1)['person_id']