所以我试图按一行中的值过滤熊猫数据框。 基本上我有一个df，其中一行包含建筑物的名称，例如。教育，K-12，办公室，教堂等。
我想根据这些值过滤一个新的数据框。 例如。我想“提取”单元格值等于“ Education，K-12”的列。 我该怎么做？
我进行了广泛搜索，但是大多数链式过滤似乎都是基于列值的。不应基于列值。
谢谢！
SAN ANTONIO, TX SAN ANTONIO, TX.1 SAN ANTONIO, TX.2 SAN ANTONIO, TX.3 \
0 Commercial Commercial Commercial Commercial
1 Fossil Fuel Fossil Fuel Fossil Fuel Fossil Fuel
2 Education, K-12 Education, K-12 Education, K-12 Education, K-12
.. ... ... ... ...
SAN ANTONIO, TX.429 SAN ANTONIO, TX.430 SAN ANTONIO, TX.431
0 Commercial Commercial Commercial
1 Electric Electric Electric
2 Office, Large Office, Large Office, Large
.. ... ... ...
[745 rows x 432 columns]>
答案 0 :(得分：1)
我的第一个想法是转置数据框
transposed = dt.T
要获得“教育，列中的K-12”
0 1 2
SAN ANTONIO, TX Commercial Fossil Fuel Education, K-12
SAN ANTONIO, TX.1 Commercial Fossil Fuel Education, K-12
SAN ANTONIO, TX.2 Commercial Fossil Fuel Office, Large
SAN ANTONIO, TX.3 Commercial Fossil Fuel Education, K-12
然后在行中搜索
transposed[ transposed[2] == 'Education, K-12' ].index
最小工作示例。
我仅使用io.StringIO来模拟内存中的文件，但是您应该使用常规文件名。
text = '''SAN ANTONIO, TX;SAN ANTONIO, TX.1;SAN ANTONIO, TX.2;SAN ANTONIO, TX.3
Commercial;Commercial;Commercial;Commercial
Fossil Fuel;Fossil Fuel;Fossil Fuel;Fossil Fuel
Education, K-12;Education, K-12;Office, Large;Education, K-12'''
import io
import pandas as pd
df = pd.read_csv(io.StringIO(text), sep=';')
print('\n--- df ---\n')
print(df)
transposed = df.T
print('\n--- transposed ---\n')
print(transposed)
print('\n--- names ---\n')
cols = transposed[ transposed[2] == 'Education, K-12' ].index
print(cols)
print('\n--- columns ---\n')
print(df[ cols ])
结果
--- df ---
SAN ANTONIO, TX SAN ANTONIO, TX.1 SAN ANTONIO, TX.2 SAN ANTONIO, TX.3
0 Commercial Commercial Commercial Commercial
1 Fossil Fuel Fossil Fuel Fossil Fuel Fossil Fuel
2 Education, K-12 Education, K-12 Office, Large Education, K-12
--- transposed ---
0 1 2
SAN ANTONIO, TX Commercial Fossil Fuel Education, K-12
SAN ANTONIO, TX.1 Commercial Fossil Fuel Education, K-12
SAN ANTONIO, TX.2 Commercial Fossil Fuel Office, Large
SAN ANTONIO, TX.3 Commercial Fossil Fuel Education, K-12
--- names ---
Index(['SAN ANTONIO, TX', 'SAN ANTONIO, TX.1', 'SAN ANTONIO, TX.3'], dtype='object')
--- columns ---
SAN ANTONIO, TX SAN ANTONIO, TX.1 SAN ANTONIO, TX.3
0 Commercial Commercial Commercial
1 Fossil Fuel Fossil Fuel Fossil Fuel
2 Education, K-12 Education, K-12 Education, K-12
答案 1 :(得分：1)
我以前从未见过这种用例。想不出一种优雅的方法，但是您可以先转置数据框，然后仅选择所需的行，然后再转回。
在下面的示例中，我将第7行设为要过滤的内容。因此，假设您要删除第7行中带有“ c”的列。因此，基本上，我们需要删除“ col2”。
>>> col1=['a','a','a','a','a','b','b','b']
>>> col2=['a','a','a','a','a','b','b','c']
>>> cols=['col1','col2']
>>> values=zip(col1,col2)
>>> import pandas as pd
>>> df=pd.DataFrame(data=values,columns=cols)
>>> df
col1 col2
0 a a
1 a a
2 a a
3 a a
4 a a
5 b b
6 b b
7 b c
>>> dft=df.T
>>> dft
0 1 2 3 4 5 6 7
col1 a a a a a b b b
col2 a a a a a b b c
>>> dff=dft[dft[7]!='c']
>>> dff
0 1 2 3 4 5 6 7
col1 a a a a a b b b
>>> dfo=dff.T
>>> dfo
col1
0 a
1 a
2 a
3 a
4 a
5 b
6 b
7 b
答案 2 :(得分：1)
测试想法后，我创建了这个
cols = df.columns[ df.iloc[2] == 'Education, K-12' ]
df[ cols ]
我只得到一行iloc[2]，所以我得到Series，并且可以将Series中的值与'Education, K-12'进行比较-这为每个项目提供True/False的值在这一行中，我可以用它来过滤列。
最小的工作示例。
我仅使用io.StringIO来模拟内存中的文件，但是您应该使用常规文件名。
text = '''SAN ANTONIO, TX;SAN ANTONIO, TX.1;SAN ANTONIO, TX.2;SAN ANTONIO, TX.3
Commercial;Commercial;Commercial;Commercial
Fossil Fuel;Fossil Fuel;Fossil Fuel;Fossil Fuel
Education, K-12;Education, K-12;Office, Large;Education, K-12'''
import io
import pandas as pd
df = pd.read_csv(io.StringIO(text), sep=';')
print('\n--- df ---\n')
print(df)
print('\n--- Series ---\n')
print( df.iloc[2] )
print('\n--- mask ---\n')
print( df.iloc[2] == 'Education, K-12' )
print('\n--- names ---\n')
cols = df.columns[ df.iloc[2] == 'Education, K-12' ]
print(cols)
print('\n--- columns ---\n')
print(df[ cols ])
结果：
--- df ---
SAN ANTONIO, TX SAN ANTONIO, TX.1 SAN ANTONIO, TX.2 SAN ANTONIO, TX.3
0 Commercial Commercial Commercial Commercial
1 Fossil Fuel Fossil Fuel Fossil Fuel Fossil Fuel
2 Education, K-12 Education, K-12 Office, Large Education, K-12
--- Series ---
SAN ANTONIO, TX Education, K-12
SAN ANTONIO, TX.1 Education, K-12
SAN ANTONIO, TX.2 Office, Large
SAN ANTONIO, TX.3 Education, K-12
Name: 2, dtype: object
--- mask ---
SAN ANTONIO, TX True
SAN ANTONIO, TX.1 True
SAN ANTONIO, TX.2 False
SAN ANTONIO, TX.3 True
Name: 2, dtype: bool
--- names ---
Index(['SAN ANTONIO, TX', 'SAN ANTONIO, TX.1', 'SAN ANTONIO, TX.3'], dtype='object')
--- columns ---
SAN ANTONIO, TX SAN ANTONIO, TX.1 SAN ANTONIO, TX.3
0 Commercial Commercial Commercial
1 Fossil Fuel Fossil Fuel Fossil Fuel
2 Education, K-12 Education, K-12 Education, K-12