Python Pandas：获取列匹配特定值的行的索引
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Python Pandas: Get index of rows which column matches certain value
提问by I want badges
Given a DataFrame with a column "BoolCol", we want to find the indexes of the DataFrame in which the values for "BoolCol" == True
给定一个带有“BoolCol”列的 DataFrame，我们想要找到“BoolCol”的值 == True 的 DataFrame 的索引
I currently have the iterating way to do it, which works perfectly:
我目前有迭代方式来做到这一点，它完美地工作：
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:
但这不是正确的熊猫方式。经过一番研究，我目前正在使用此代码：
df[df['BoolCol'] == True].index.tolist()
This one gives me a list of indexes, but they dont match, when I check them by doing:
这个给了我一个索引列表，但当我通过以下方式检查它们时，它们不匹配：
df.iloc[i]['BoolCol']
The result is actually False!!
结果居然是假的！！
Which would be the correct Pandas way to do this?
哪个是 Pandas 的正确方法？
采纳答案by unutbu
df.iloc[i]returns the ithrow of df. idoes not refer to the index label, iis a 0-based index.
df.iloc[i]返回 的ith行df。i不引用索引标签，i是一个从0开始的索引。
In contrast, the attribute indexreturns actual index labels, not numeric row-indices:
相反，该属性index返回实际索引标签，而不是数字行索引：
df.index[df['BoolCol'] == True].tolist()
or equivalently,
或等效地，
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:
通过使用具有不等于行的数字位置的非默认索引的 DataFrame，您可以非常清楚地看到差异：
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,
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')
then you can select the rows using locinstead of iloc:
那么您可以使用loc而不是选择行iloc：
In [58]: df.loc[idx]
Out[58]:
BoolCol
10 True
40 True
50 True
[3 rows x 1 columns]
Note that loccan also accept boolean arrays:
请注意，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]
If you have a boolean array, mask, and need ordinal index values, you can compute them using np.flatnonzero:
如果您有一个布尔数组 ,mask并且需要序数索引值，您可以使用np.flatnonzero以下方法计算它们：
In [110]: np.flatnonzero(df['BoolCol'])
Out[112]: array([0, 3, 4])
Use df.ilocto select rows by ordinal index:
用于df.iloc按序数索引选择行：
In [113]: df.iloc[np.flatnonzero(df['BoolCol'])]
Out[113]:
BoolCol
10 True
40 True
50 True
回答by Surya
Can be done using numpy where() function:
可以使用 numpy where() 函数完成：
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
Though you don't always need index for a match, but incase if you need:
虽然您并不总是需要匹配的索引，但如果您需要：
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by YOBEN_S
First you may check querywhen the target column is type bool(PS: about how to use it please check link)
首先您可以检查query目标列何时输入bool（PS：有关如何使用它请检查链接）
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 .
在我们通过布尔列过滤原始 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')
Also pandas have nonzero, we just select the positionof Truerow and using it slice the DataFrameor index
也有大熊猫nonzero，我们只需选择位置的True一排并用它切片DataFrame或index
df.index[df.BoolCol.nonzero()[0]]
Out[128]: Int64Index([10, 40, 50], dtype='int64')
回答by Ben Druitt
Simple way is to reset the index of the DataFrame prior to filtering:
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()
Bit hacky, but it's quick!
有点hacky，但它很快！
回答by Carson
I extended this question that is how to gets the row, columnand valueof all matches value?
我扩展了这个问题，即如何获取row,column以及value所有匹配项的值？
here is solution:
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]
Output:
输出：
0 1 2
0 cat dog
1 goldfish kitten
2 Puppy hamster mouse
row col name
0 1 dog
2 0 Puppy
回答by mbh86
If you want to use your dataframe object only once, use:
如果您只想使用数据框对象一次，请使用：
df['BoolCol'].loc[lambda x: x==True].index
