blocks|key|4660755|text|可以使用numpy+where()函数来完成：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4660756|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|code-block|syntax|javascript|4660757|虽然你并不总是需要索引来匹配，但是如果你需要：|4660758|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']|4660759|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Can be done using numpy where() function:

<pre><code>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
</code></pre>

Though you don't always need index for a match, but incase if you need:

<pre><code>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']
</code></pre>

blocks|key|4660777|text|简单的方法是在过滤前重置DataFrame的索引：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4660778|df_reset+=+df.reset_index()
df_reset[df_reset['BoolCol']].index.tolist()|code-block|syntax|javascript|4660779|有点老生常谈，但很快！|4660780|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Simple way is to reset the index of the DataFrame prior to filtering:

<pre><code>df_reset = df.reset_index()
df_reset[df_reset['BoolCol']].index.tolist()
</code></pre>

Bit hacky, but it's quick!

blocks|key|154366|text|我扩展了这个问题，如何获得所有匹配值的row，column和value？|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|154367|以下是解决方案：|154368|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]|code-block|syntax|javascript|154369|输出：|154370|++++++++++0++++++++1+++++++2
0+++++++cat++++++dog++++++++
1++goldfish+++++++++++kitten
2+++++Puppy++hamster+++mouse


row++col++++++++name
0++++1+++++++++++dog
2++++0+++++++++Puppy|154371|entityMap^0|J|3|N|6|U|5|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|11|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|R|$]]

I extended this question that is how to gets the <code>row</code>, <code>column</code>and <code>value</code> of all matches value?

here is solution:

<pre><code>import pandas as pd
import numpy as np


def search_coordinate(df_data: pd.DataFrame, search_set: set) -&gt; 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':&lt;4} {'col':&lt;4} {'name':&gt;10}")
 [print(f"{row:&lt;4} {col:&lt;4} {name:&gt;10}") for row, col, name in result_list]
</code></pre>

Output:

<pre><code> 0 1 2
0 cat dog 
1 goldfish kitten
2 Puppy hamster mouse


row col name
0 1 dog
2 0 Puppy
</code></pre>

blocks|key|4660828|text|如果您只想使用dataframe对象一次，请使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4660829|df['BoolCol'].loc[lambda+x:+x==True].index|code-block|syntax|javascript|4660830|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you want to use your dataframe object only once, use:

<pre><code>df['BoolCol'].loc[lambda x: x==True].index
</code></pre>

blocks|key|4660868|text|对于我们感兴趣的已知索引候选，一种不检查整个列的更快方法可以这样完成：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4660869|np.array(index_slice)[np.where(df.loc[index_slice]['column_name']+>=+threshold)[0]]|code-block|syntax|javascript|4660870|完整比较：|4660871|import+pandas+as+pd
import+numpy+as+np

index_slice+=+list(range(50,150))+#+know+index+location+for+our+inteterest
data+=+np.zeros(10000)
data[(index_slice)]+=+np.random.random(len(index_slice))

df+=+pd.DataFrame(
++++{'column_name':+data},
)

threshold+=+0.5|4660872|%25%25timeit
np.array(index_slice)[np.where(df.loc[index_slice]['column_name']+>=+threshold)[0]]
#+600+µs+±+1.21+µs+per+loop+(mean+±+std.+dev.+of+7+runs,+1000+loops+each)

%25%25timeit
[i+for+i+in+index_slice+if+i+in+df.index[df['column_name']+>=+threshold].tolist()]
#+22.5+ms+±+29.1+µs+per+loop+(mean+±+std.+dev.+of+7+runs,+10+loops+each)|4660873|它的工作方式如下：|4660874|#+generate+Boolean+satisfy+condition+only+in+sliced+column
df.loc[index_slice]['column_name']+>=+threshold

#+convert+Boolean+to+index,+but+start+from+0+and+increment+by+1
np.where(...)[0]

#+list+of+index+to+be+sliced
np.array(index_slice)[...]|4660875|注意:需要注意的是，由于np.where(...)[0]索引start+from+0+and+increment+by+1，df.index不能替代np.where(...)[0]，但您可以创建类似df.index[index_slice]的内容。我认为，如果你只用少量的行数来做一次，这是不值得的。|offset|length|style|CODE|4660876|entityMap^0|0|0|0|0|0|0|0|C|G|U|V|1Q|8|22|G|2R|L|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|D|7|12|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|13|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|15|8|@$S|16|T|17|U|V]|$S|18|T|19|U|V]|$S|1A|T|1B|U|V]|$S|1C|T|1D|U|V]|$S|1E|T|1F|U|V]]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]]|X|$]]

For known index candidate that we interested, a faster way by not checking the whole column can be done like this:
<pre class="lang-py prettyprint-override"><code>np.array(index_slice)[np.where(df.loc[index_slice]['column_name'] &gt;= threshold)[0]]
</code></pre>
Full comparison:
<pre class="lang-py prettyprint-override"><code>import pandas as pd
import numpy as np

index_slice = list(range(50,150)) # know index location for our inteterest
data = np.zeros(10000)
data[(index_slice)] = np.random.random(len(index_slice))

df = pd.DataFrame(
 {'column_name': data},
)

threshold = 0.5

</code></pre>
<pre class="lang-py prettyprint-override"><code>%%timeit
np.array(index_slice)[np.where(df.loc[index_slice]['column_name'] &gt;= threshold)[0]]
# 600 µs ± 1.21 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%%timeit
[i for i in index_slice if i in df.index[df['column_name'] &gt;= threshold].tolist()]
# 22.5 ms ± 29.1 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
</code></pre>
The way it works is like this:
<pre class="lang-py prettyprint-override"><code># generate Boolean satisfy condition only in sliced column
df.loc[index_slice]['column_name'] &gt;= threshold

# convert Boolean to index, but start from 0 and increment by 1
np.where(...)[0]

# list of index to be sliced
np.array(index_slice)[...]
</code></pre>
Note:
It needs to be noted that <code>np.array(index_slice)</code> can't be substituted by <code>df.index</code> due to <code>np.where(...)[0]</code> indexing <code>start from 0 and increment by 1</code>, but you can make something like <code>df.index[index_slice]</code>. And I think this is not worth the hassle if you just do it one time with small number of rows.

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:

<pre><code>for i in range(100,3000):
 if df.iloc[i]['BoolCol']== True:
 print i,df.iloc[i]['BoolCol']
</code></pre>

But this is not the correct panda's way to do it.
After some research, I am currently using this code:

<pre><code>df[df['BoolCol'] == True].index.tolist()
</code></pre>

This one gives me a list of indexes, but they dont match, when I check them by doing:

<pre><code>df.iloc[i]['BoolCol']
</code></pre>

The result is actually False!!

Which would be the correct Pandas way to do this?

Python Pandas: Get index of rows which column matches certain value

给定一个具有列"BoolCol“的DataFrame，我们希望找到其中"BoolCol”的值为真的DataFrame的索引我目前有一种迭代的方式来做这件事，它工作得很好：for i in range(100,3000):    if df.iloc[i]['BoolCol']== True:         print...

问Python Pandas:获取列与特定值匹配的行的索引
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