[英]Check if column value is in other columns in pandas
I have the following dataframe in pandas 我在熊猫中有以下数据帧
target A B C
0 cat bridge cat brush
1 brush dog cat shoe
2 bridge cat shoe bridge
How do I test whether df.target
is in any of the columns ['A','B','C', etc.]
, where there are many columns to check? 如何测试df.target
是否在任何列['A','B','C', etc.]
,哪些列要检查?
I have tried merging A,B and C into a string to use df.abcstring.str.contains(df.target)
but this does not work. 我尝试将A,B和C合并为字符串以使用df.abcstring.str.contains(df.target)
但这不起作用。
You can use drop
, isin
and any
. 你可以使用drop
, isin
和any
。
drop
the target
column to have a df with your A
, B
, C
columns only drop
target
列,使其仅包含A
, B
, C
列的df isin
the target column 检查值isin
目标列 any
hits are present 并检查是否有any
命中 That's it. 而已。
df["exists"] = df.drop("target", 1).isin(df["target"]).any(1)
print(df)
target A B C exists
0 cat bridge cat brush True
1 brush dog cat shoe False
2 bridge cat shoe bridge True
OneHotEncoder approach: OneHotEncoder方法:
In [165]: x = pd.get_dummies(df.drop('target',1), prefix='', prefix_sep='')
In [166]: x
Out[166]:
bridge cat dog cat shoe bridge brush shoe
0 1 0 0 1 0 0 1 0
1 0 0 1 1 0 0 0 1
2 0 1 0 0 1 1 0 0
In [167]: x[df['target']].eq(1).any(1)
Out[167]:
0 True
1 True
2 True
dtype: bool
Explanation: 说明:
In [168]: x[df['target']]
Out[168]:
cat cat brush bridge bridge
0 0 1 1 1 0
1 0 1 0 0 0
2 1 0 0 0 1
You can use eq
, for drop column pop
if neech check by rows: 如果neech按行检查,你可以使用eq
,drop drop pop
:
mask = df.eq(df.pop('target'), axis=0)
print (mask)
A B C
0 False True False
1 False False False
2 False False True
And then if need check at least one True
add any
: 然后如果需要检查至少一个True
添加any
:
mask = df.eq(df.pop('target'), axis=0).any(axis=1)
print (mask)
0 True
1 False
2 True
dtype: bool
df['new'] = df.eq(df.pop('target'), axis=0).any(axis=1)
print (df)
A B C new
0 bridge cat brush True
1 dog cat shoe False
2 cat shoe bridge True
But if need check all values in column use isin
: 但如果需要检查列使用中的所有值isin
:
mask = df.isin(df.pop('target').values.tolist())
print (mask)
A B C
0 True True True
1 False True False
2 True False True
And if want check if all values are True
add all
: 如果想检查所有值是否为True
添加all
:
df['new'] = df.isin(df.pop('target').values.tolist()).all(axis=1)
print (df)
A B C new
0 bridge cat brush True
1 dog cat shoe False
2 cat shoe bridge False
you can use apply a function for each row that counts the number of value that match the value in the 'target' column: 您可以使用为每行计算一个函数,该函数计算与“目标”列中的值匹配的值的数量:
df["exist"] = df.apply(lambda row:row.value_counts()[row['target']] > 1 , axis=1)
for a dataframe that looks like: 对于看起来像这样的数据框:
b c target
0 3 a a
1 3 4 2
2 3 4 2
3 3 4 2
4 3 4 4
the output will be: 输出将是:
b c target exist
0 3 a a True
1 3 4 2 False
2 3 4 2 False
3 3 4 2 False
4 3 4 4 True
Another approach using index difference method: 另一种使用索引差异法的方法:
matches = df[df.columns.difference(['target'])].eq(df['target'], axis = 0)
# A B C
#0 False True False
#1 False False False
#2 False False True
# Check if at least one match:
matches.any(axis = 1)
#Out[30]:
#0 True
#1 False
#2 True
In case you wanted to see which columns meet the target, here is a possible solution: 如果您想查看哪些列符合目标,这是一个可能的解决方案:
matches.apply(lambda x: ", ".join(x.index[np.where(x.tolist())]), axis = 1)
Out[53]:
0 B
1
2 C
dtype: object
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