问题:Python Pandas:获取列匹配特定值的行的索引
给定一个带有“ BoolCol”列的DataFrame,我们要查找其中“ BoolCol” == True的值的DataFrame索引
我目前有迭代的方式来做,很完美:
for i in range(100,3000):
if df.iloc[i]['BoolCol']== True:
print i,df.iloc[i]['BoolCol']
但这不是正确的熊猫方法。经过研究,我目前正在使用以下代码:
df[df['BoolCol'] == True].index.tolist()
这给了我一份索引列表,但是当我通过以下方法检查它们时,它们不匹配:
df.iloc[i]['BoolCol']
结果实际上是错误的!
哪一种是正确的Pandas方法?
Given a DataFrame with a column “BoolCol”, we want to find the indexes of the DataFrame in which the values for “BoolCol” == True
I currently have the iterating way to do it, which works perfectly:
for i in range(100,3000):
if df.iloc[i]['BoolCol']== True:
print i,df.iloc[i]['BoolCol']
But this is not the correct panda’s way to do it.
After some research, I am currently using this code:
df[df['BoolCol'] == True].index.tolist()
This one gives me a list of indexes, but they dont match, when I check them by doing:
df.iloc[i]['BoolCol']
The result is actually False!!
Which would be the correct Pandas way to do this?
回答 0
df.iloc[i]
返回的ith
行df
。i
不引用索引标签,i
是基于0的索引。
相反,该属性index
返回实际的索引标签,而不是数字的行索引:
df.index[df['BoolCol'] == True].tolist()
或等效地,
df.index[df['BoolCol']].tolist()
通过使用具有非默认索引的DataFrame玩,可以很清楚地看到差异,该索引与行的数字位置不相等:
df = pd.DataFrame({'BoolCol': [True, False, False, True, True]},
index=[10,20,30,40,50])
In [53]: df
Out[53]:
BoolCol
10 True
20 False
30 False
40 True
50 True
[5 rows x 1 columns]
In [54]: df.index[df['BoolCol']].tolist()
Out[54]: [10, 40, 50]
如果要使用索引,
In [56]: idx = df.index[df['BoolCol']]
In [57]: idx
Out[57]: Int64Index([10, 40, 50], dtype='int64')
那么您可以使用loc
代替来选择行iloc
:
In [58]: df.loc[idx]
Out[58]:
BoolCol
10 True
40 True
50 True
[3 rows x 1 columns]
注意,loc
也可以接受布尔数组:
In [55]: df.loc[df['BoolCol']]
Out[55]:
BoolCol
10 True
40 True
50 True
[3 rows x 1 columns]
如果您有一个布尔数组,mask
并且需要序数索引值,则可以使用进行计算np.flatnonzero
:
In [110]: np.flatnonzero(df['BoolCol'])
Out[112]: array([0, 3, 4])
用于df.iloc
按顺序索引选择行:
In [113]: df.iloc[np.flatnonzero(df['BoolCol'])]
Out[113]:
BoolCol
10 True
40 True
50 True
df.iloc[i]
returns the ith
row of df
. i
does not refer to the index label, i
is a 0-based index.
In contrast, the attribute index
returns actual index labels, not numeric row-indices:
df.index[df['BoolCol'] == True].tolist()
or equivalently,
df.index[df['BoolCol']].tolist()
You can see the difference quite clearly by playing with a DataFrame with
a non-default index that does not equal to the row’s numerical position:
df = pd.DataFrame({'BoolCol': [True, False, False, True, True]},
index=[10,20,30,40,50])
In [53]: df
Out[53]:
BoolCol
10 True
20 False
30 False
40 True
50 True
[5 rows x 1 columns]
In [54]: df.index[df['BoolCol']].tolist()
Out[54]: [10, 40, 50]
If you want to use the index,
In [56]: idx = df.index[df['BoolCol']]
In [57]: idx
Out[57]: Int64Index([10, 40, 50], dtype='int64')
then you can select the rows using loc
instead of iloc
:
In [58]: df.loc[idx]
Out[58]:
BoolCol
10 True
40 True
50 True
[3 rows x 1 columns]
Note that loc
can also accept boolean arrays:
In [55]: df.loc[df['BoolCol']]
Out[55]:
BoolCol
10 True
40 True
50 True
[3 rows x 1 columns]
If you have a boolean array, mask
, and need ordinal index values, you can compute them using np.flatnonzero
:
In [110]: np.flatnonzero(df['BoolCol'])
Out[112]: array([0, 3, 4])
Use df.iloc
to select rows by ordinal index:
In [113]: df.iloc[np.flatnonzero(df['BoolCol'])]
Out[113]:
BoolCol
10 True
40 True
50 True
回答 1
可以使用numpy where()函数来完成:
import pandas as pd
import numpy as np
In [716]: df = pd.DataFrame({"gene_name": ['SLC45A1', 'NECAP2', 'CLIC4', 'ADC', 'AGBL4'] , "BoolCol": [False, True, False, True, True] },
index=list("abcde"))
In [717]: df
Out[717]:
BoolCol gene_name
a False SLC45A1
b True NECAP2
c False CLIC4
d True ADC
e True AGBL4
In [718]: np.where(df["BoolCol"] == True)
Out[718]: (array([1, 3, 4]),)
In [719]: select_indices = list(np.where(df["BoolCol"] == True)[0])
In [720]: df.iloc[select_indices]
Out[720]:
BoolCol gene_name
b True NECAP2
d True ADC
e True AGBL4
虽然您并不总是需要索引来进行匹配,但是如果需要的话:
In [796]: df.iloc[select_indices].index
Out[796]: Index([u'b', u'd', u'e'], dtype='object')
In [797]: df.iloc[select_indices].index.tolist()
Out[797]: ['b', 'd', 'e']
Can be done using numpy where() function:
import pandas as pd
import numpy as np
In [716]: df = pd.DataFrame({"gene_name": ['SLC45A1', 'NECAP2', 'CLIC4', 'ADC', 'AGBL4'] , "BoolCol": [False, True, False, True, True] },
index=list("abcde"))
In [717]: df
Out[717]:
BoolCol gene_name
a False SLC45A1
b True NECAP2
c False CLIC4
d True ADC
e True AGBL4
In [718]: np.where(df["BoolCol"] == True)
Out[718]: (array([1, 3, 4]),)
In [719]: select_indices = list(np.where(df["BoolCol"] == True)[0])
In [720]: df.iloc[select_indices]
Out[720]:
BoolCol gene_name
b True NECAP2
d True ADC
e True AGBL4
Though you don’t always need index for a match, but incase if you need:
In [796]: df.iloc[select_indices].index
Out[796]: Index([u'b', u'd', u'e'], dtype='object')
In [797]: df.iloc[select_indices].index.tolist()
Out[797]: ['b', 'd', 'e']
回答 2
一种简单的方法是在过滤之前重置DataFrame的索引:
df_reset = df.reset_index()
df_reset[df_reset['BoolCol']].index.tolist()
有点hacky,但是很快!
Simple way is to reset the index of the DataFrame prior to filtering:
df_reset = df.reset_index()
df_reset[df_reset['BoolCol']].index.tolist()
Bit hacky, but it’s quick!
回答 3
首先,您可以检查query
目标列的类型bool
(PS:关于如何使用它,请检查链接)
df.query('BoolCol')
Out[123]:
BoolCol
10 True
40 True
50 True
在通过Boolean列过滤原始df之后,我们可以选择索引。
df=df.query('BoolCol')
df.index
Out[125]: Int64Index([10, 40, 50], dtype='int64')
大熊猫也有nonzero
,我们只需选择行的位置,True
然后使用它对DataFrame
或index
df.index[df.BoolCol.nonzero()[0]]
Out[128]: Int64Index([10, 40, 50], dtype='int64')
First you may check query
when the target column is type bool
(PS: about how to use it please check link )
df.query('BoolCol')
Out[123]:
BoolCol
10 True
40 True
50 True
After we filter the original df by the Boolean column we can pick the index .
df=df.query('BoolCol')
df.index
Out[125]: Int64Index([10, 40, 50], dtype='int64')
Also pandas have nonzero
, we just select the position of True
row and using it slice the DataFrame
or index
df.index[df.BoolCol.nonzero()[0]]
Out[128]: Int64Index([10, 40, 50], dtype='int64')
回答 4
如果只想使用一次数据框对象,请使用:
df['BoolCol'].loc[lambda x: x==True].index
If you want to use your dataframe object only once, use:
df['BoolCol'].loc[lambda x: x==True].index
回答 5
我扩展这个问题是如何获取row
,column
并且value
所有的比赛价值?
这是解决方案:
import pandas as pd
import numpy as np
def search_coordinate(df_data: pd.DataFrame, search_set: set) -> list:
nda_values = df_data.values
tuple_index = np.where(np.isin(nda_values, [e for e in search_set]))
return [(row, col, nda_values[row][col]) for row, col in zip(tuple_index[0], tuple_index[1])]
if __name__ == '__main__':
test_datas = [['cat', 'dog', ''],
['goldfish', '', 'kitten'],
['Puppy', 'hamster', 'mouse']
]
df_data = pd.DataFrame(test_datas)
print(df_data)
result_list = search_coordinate(df_data, {'dog', 'Puppy'})
print(f"\n\n{'row':<4} {'col':<4} {'name':>10}")
[print(f"{row:<4} {col:<4} {name:>10}") for row, col, name in result_list]
输出:
0 1 2
0 cat dog
1 goldfish kitten
2 Puppy hamster mouse
row col name
0 1 dog
2 0 Puppy
I extended this question that is how to gets the row
, column
and value
of all matches value?
here is solution:
import pandas as pd
import numpy as np
def search_coordinate(df_data: pd.DataFrame, search_set: set) -> list:
nda_values = df_data.values
tuple_index = np.where(np.isin(nda_values, [e for e in search_set]))
return [(row, col, nda_values[row][col]) for row, col in zip(tuple_index[0], tuple_index[1])]
if __name__ == '__main__':
test_datas = [['cat', 'dog', ''],
['goldfish', '', 'kitten'],
['Puppy', 'hamster', 'mouse']
]
df_data = pd.DataFrame(test_datas)
print(df_data)
result_list = search_coordinate(df_data, {'dog', 'Puppy'})
print(f"\n\n{'row':<4} {'col':<4} {'name':>10}")
[print(f"{row:<4} {col:<4} {name:>10}") for row, col, name in result_list]
Output:
0 1 2
0 cat dog
1 goldfish kitten
2 Puppy hamster mouse
row col name
0 1 dog
2 0 Puppy