blocks|key|386080|text|使用DataFrame.drop并向其传递一系列索引标签：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|386081|In+[65]:+df
Out[65]:+
+++++++one++two
one++++++1++++4
two++++++2++++3
three++++3++++2
four+++++4++++1


In+[66]:+df.drop(df.index[[1,3]])
Out[66]:+
+++++++one++two
one++++++1++++4
three++++3++++2|code-block|syntax|javascript|386082|entityMap|0|LINK|mutability|MUTABLE|url|http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.drop.html^0|2|E|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Use <a href="http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.drop.html">DataFrame.drop</a> and pass it a Series of index labels:

<pre><code>In [65]: df
Out[65]: 
 one two
one 1 4
two 2 3
three 3 2
four 4 1


In [66]: df.drop(df.index[[1,3]])
Out[66]: 
 one two
one 1 4
three 3 2
</code></pre>

blocks|key|386099|text|请注意，当您想要插入行时，使用"inplace“命令可能很重要。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|386100|df.drop(df.index[[1,3]],+inplace=True)|code-block|syntax|javascript|386101|因为您的原始问题没有返回任何内容，所以应该使用此命令。http://pandas.pydata.org/pandas-docs/version/0.17.0/generated/pandas.DataFrame.drop.html|offset|length|386102|entityMap|0|LINK|mutability|MUTABLE|url|http://pandas.pydata.org/pandas-docs/version/0.17.0/generated/pandas.DataFrame.drop.html^0|0|0|R|2G|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

Note that it may be important to use the "inplace" command when you want to do the drop in line. 

<pre><code>df.drop(df.index[[1,3]], inplace=True)
</code></pre>

Because your original question is not returning anything, this command should be used.
<a href="http://pandas.pydata.org/pandas-docs/version/0.17.0/generated/pandas.DataFrame.drop.html">http://pandas.pydata.org/pandas-docs/version/0.17.0/generated/pandas.DataFrame.drop.html</a>

blocks|key|168130|text|在对@theodros-zelleke的回答的评论中，@j-jones询问了如果索引不是唯一的该怎么办。我不得不处理这样的情况。我所做的是在调用drop()之前重命名索引中的重复项：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|168131|dropped_indexes+=+<determine-indexes-to-drop>
df.index+=+rename_duplicates(df.index)
df.drop(df.index[dropped_indexes],+inplace=True)|code-block|syntax|javascript|168132|其中rename_duplicates()是我定义的一个函数，它遍历了索引的元素并重命名了副本。我使用了与pd.read_csv()在列上使用的相同的重命名模式，即"%25s.%25d"+%25+(name,+count)，其中name是行的名称，count是它之前出现的次数。|168133|entityMap^0|20|6|0|0|2|J|1H|D|2A|N|30|4|3A|5|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|N|$]]

In a comment to @theodros-zelleke's answer, @j-jones asked about what to do if the index is not unique. I had to deal with such a situation. What I did was to rename the duplicates in the index before I called <code>drop()</code>, a la:

<pre><code>dropped_indexes = &lt;determine-indexes-to-drop&gt;
df.index = rename_duplicates(df.index)
df.drop(df.index[dropped_indexes], inplace=True)
</code></pre>

where <code>rename_duplicates()</code> is a function I defined that went through the elements of index and renamed the duplicates. I used the same renaming pattern as <code>pd.read_csv()</code> uses on columns, i.e., <code>"%s.%d" % (name, count)</code>, where <code>name</code> is the name of the row and <code>count</code> is how many times it has occurred previously.

blocks|key|386228|text|如果DataFrame很大，而且要删除的行数也很大，那么简单的drop+by+index+df.drop(df.index[])就会花费太多时间。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|386229|在我的例子中，我使用100M+rows+x+3+cols创建了一个浮点数的多索引DataFrame，并且我需要从其中删除10k行。我发现的最快的方法是，与直觉相反的是，对其余的行执行take。|386230|假设indexes_to_drop是要删除的位置索引的数组(在问题中为[1,+2,+4])。|386231|indexes_to_keep+=+set(range(df.shape[0]))+-+set(indexes_to_drop)
df_sliced+=+df.take(list(indexes_to_keep))|code-block|syntax|javascript|386232|在我的例子中，这占用了20.5s，而简单的df.drop占用了5min+27s并消耗了大量内存。生成的DataFrame是相同的。|386233|entityMap^0|19|J|0|A|I|1O|3|2J|4|0|2|F|Z|9|0|0|B|5|L|7|V|8|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@$9|W|A|X|B|C]|$9|Y|A|Z|B|C]|$9|10|A|11|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@]|E|$]]|$1|J|3|K|5|L|7|17|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|R|$]]

If the DataFrame is huge, and the number of rows to drop is large as well, then simple drop by index <code>df.drop(df.index[])</code> takes too much time. 

In my case, I have a multi-indexed DataFrame of floats with <code>100M rows x 3 cols</code>, and I need to remove <code>10k</code> rows from it. The fastest method I found is, quite counterintuitively, to <code>take</code> the remaining rows.

Let <code>indexes_to_drop</code> be an array of positional indexes to drop (<code>[1, 2, 4]</code> in the question).

<pre><code>indexes_to_keep = set(range(df.shape[0])) - set(indexes_to_drop)
df_sliced = df.take(list(indexes_to_keep))
</code></pre>

In my case this took <code>20.5s</code>, while the simple <code>df.drop</code> took <code>5min 27s</code> and consumed a lot of memory. The resulting DataFrame is the same.

blocks|key|386262|text|如果我想删除一行，假设它有索引x，我将执行以下操作：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|386263|df+=+df[df.index+!=+x]|code-block|syntax|javascript|386264|如果我想删除多个索引(假设这些索引在列表unwanted_indices中)，我会这样做：|386265|desired_indices+=+[i+for+i+in+len(df.index)+if+i+not+in+unwanted_indices]
desired_df+=+df.iloc[desired_indices]|386266|entityMap^0|F|1|0|0|K|G|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

If I want to drop a row which has let's say index <code>x</code>, I would do the following:

<pre><code>df = df[df.index != x]
</code></pre>

If I would want to drop multiple indices (say these indices are in the list <code>unwanted_indices</code>), I would do:

<pre><code>desired_indices = [i for i in len(df.index) if i not in unwanted_indices]
desired_df = df.iloc[desired_indices]
</code></pre>

blocks|key|386291|text|这里有一个具体的例子，我想展示一下。假设在某些行中有许多重复的条目。如果您有字符串条目，那么可以很容易地使用string方法来查找要删除的所有索引。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|386292|ind_drop+=+df[df['column_of_strings'].apply(lambda+x:+x.startswith('Keyword'))].index|code-block|syntax|javascript|386293|现在使用它们的索引删除这些行|386294|new_df+=+df.drop(ind_drop)|386295|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|$]]

Here is a bit specific example, I would like to show. Say you have many duplicate entries in some of your rows. If you have string entries you could easily use string methods to find all indexes to drop. 

<pre><code>ind_drop = df[df['column_of_strings'].apply(lambda x: x.startswith('Keyword'))].index
</code></pre>

And now to drop those rows using their indexes 

<pre><code>new_df = df.drop(ind_drop)
</code></pre>

blocks|key|386327|text|如上所述从布尔值确定索引，例如|type|unstyled|depth|inlineStyleRanges|entityRanges|data|386328|df[df['column'].isin(values)].index|code-block|syntax|javascript|386329|可以比使用此方法确定索引占用更多的内存|386330|pd.Index(np.where(df['column'].isin(values))[0])|386331|像这样应用|386332|df.drop(pd.Index(np.where(df['column'].isin(values))[0]),+inplace+=+True)|386333|这种方法在处理大数据帧和有限内存时很有用。|386334|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Determining the index from the boolean as described above e.g.

<pre><code>df[df['column'].isin(values)].index
</code></pre>

can be more memory intensive than determining the index using this method

<pre><code>pd.Index(np.where(df['column'].isin(values))[0])
</code></pre>

applied like so

<pre><code>df.drop(pd.Index(np.where(df['column'].isin(values))[0]), inplace = True)
</code></pre>

This method is useful when dealing with large dataframes and limited memory.

blocks|key|168325|text|仅使用索引参数删除行：-|type|unstyled|depth|inlineStyleRanges|entityRanges|data|168326|df.drop(index+=+2,+inplace+=+True)|code-block|syntax|javascript|168327|对于多行：-|168328|df.drop(index=[1,3],+inplace+=+True)|168329|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|$]]

Use only the Index arg to drop row:-

<pre><code>df.drop(index = 2, inplace = True)
</code></pre>

For multiple rows:-

<pre><code>df.drop(index=[1,3], inplace = True)
</code></pre>

blocks|key|386448|text|要删除索引为1、2、4的行，可以使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|386449|df[~df.index.isin([1,+2,+4])]|code-block|syntax|javascript|386450|代字号运算符~对方法isin的结果求反。另一种选择是删除索引：|offset|length|style|CODE|386451|df.loc[df.index.drop([1,+2,+4])]|386452|entityMap^0|0|0|6|1|A|4|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|P|$]]

To drop rows with indices 1, 2, 4 you can use:
<pre><code>df[~df.index.isin([1, 2, 4])]
</code></pre>
The tilde operator <code>~</code> negates the result of the method <code>isin</code>. Another option is to drop indices:
<pre><code>df.loc[df.index.drop([1, 2, 4])]
</code></pre>

blocks|key|168434|text|正如Dennis+Golomazov's+answer建议的那样，使用drop删除行。您可以选择保留行。假设您有一个名为indices_to_drop的要删除的行索引列表。您可以将其转换为掩码，如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|168435|mask+=+np.ones(len(df),+bool)
mask[indices_to_drop]+=+False|code-block|syntax|javascript|168436|您可以直接使用此索引：|168437|df_new+=+df.iloc[mask]|168438|这种方法的好处是，mask可以来自任何来源:它可以是包含许多列的条件，也可以是其他条件。|168439|真正好的事情是，您实际上根本不需要原始DataFrame的索引，所以索引是否唯一并不重要。|168440|当然，缺点是您不能使用这种方法进行原地投放。|168441|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/43421391/2988730^0|Z|4|1O|F|2|P|0|0|0|0|0|9|4|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@$9|17|A|18|1|19]]|E|$]]|$1|F|3|G|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1B|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1C|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1D|8|@$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1G|8|@]|D|@]|E|$]]|$1|S|3|T|5|6|7|1H|8|@]|D|@]|E|$]]|$1|U|3|-4|5|6|7|1I|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11]]]]

As <a href="https://stackoverflow.com/a/43421391/2988730">Dennis Golomazov's answer</a> suggests, using <code>drop</code> to drop rows. You can select to keep rows instead. Let's say you have a list of row indices to drop called <code>indices_to_drop</code>. You can convert it to a mask as follows:
<pre><code>mask = np.ones(len(df), bool)
mask[indices_to_drop] = False
</code></pre>
You can use this index directly:
<pre><code>df_new = df.iloc[mask]
</code></pre>
The nice thing about this method is that <code>mask</code> can come from any source: it can be a condition involving many columns, or something else.
The really nice thing is, you really don't need the index of the original DataFrame at all, so it doesn't matter if the index is unique or not.
The disadvantage is of course that you can't do the drop in-place with this method.

I have a dataframe df :

<pre><code>&gt;&gt;&gt; df
 sales discount net_sales cogs
STK_ID RPT_Date 
600141 20060331 2.709 NaN 2.709 2.245
 20060630 6.590 NaN 6.590 5.291
 20060930 10.103 NaN 10.103 7.981
 20061231 15.915 NaN 15.915 12.686
 20070331 3.196 NaN 3.196 2.710
 20070630 7.907 NaN 7.907 6.459
</code></pre>

Then I want to drop rows with certain sequence numbers which indicated in a list, suppose here is <code>[1,2,4],</code> then left:

<pre><code> sales discount net_sales cogs
STK_ID RPT_Date 
600141 20060331 2.709 NaN 2.709 2.245
 20061231 15.915 NaN 15.915 12.686
 20070630 7.907 NaN 7.907 6.459
</code></pre>

How or what function can do that ?

How to drop a list of rows from Pandas dataframe?

我有一个数据帧df：>>> df                  sales  discount  net_sales    cogsSTK_ID RPT_Date                                     600141 20060331   2.709       NaN      2...

问如何从Pandas dataframe中删除行的列表？
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