blocks|key|220451|text|使用Series.map怎么样？|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|220452|import+pandas+as+pd

df+=+pd.DataFrame({'x':+['good',+'bad',+'bad',+'good']})
print(df)
df['x']+=+df.x.map({'good':+'bad',+'bad':+'good'})
print(df)|code-block|syntax|javascript|220453|++++++x+++++++++++x
0++good+++++0+++bad
1+++bad++→++1++good
2+++bad+++++2++good
3++good+++++3+++bad|220454|220455|对于问题的编辑部分，可以将Series.map和Series.where组合起来。|220456|import+pandas+as+pd

df+=+pd.DataFrame({'x':+['good',+'bad',+'bad',+'good'],+'y':+[100,+0,+100,+0]})
print(df)
df['x']+=+df.x.map({'good':+'bad',+'bad':+'good'}).where(df.y+>=+100,+df.x)
print(df)|220457|其思想是全局交换值(使用map)，然后返回(使用where)到条件df.y+>=+100为False的原始x。|220458|或者，您可以使用DataFrame.apply|220459|swap+=+{'good':+'bad',+'bad':+'good'}
df['x']+=+df.apply(lambda+row:+swap[row.x]+if+row.y+>=+100+else+row.x,+axis=1)|220460|++++++x++++y+++++++++++x++++y
0++good++100+++++0+++bad++100
1+++bad++++0++→++1+++bad++++0
2+++bad++100+++++2++good++100
3++good++++0+++++3++good++++0|220461|entityMap|0|LINK|mutability|MUTABLE|url|https://pandas.pydata.org/docs/reference/api/pandas.Series.map.html#pandas.Series.map^0|2|A|2|A|0|0|0|0|0|D|A|O|C|0|0|C|3|O|5|X|B|19|5|1H|1|0|8|F|0|0|0^^$0|@$1|2|3|4|5|6|7|17|8|@$9|18|A|19|B|C]]|D|@$9|1A|A|1B|1|1C]]|E|$]]|$1|F|3|G|5|H|7|1D|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|H|7|1E|8|@]|D|@]|E|$I|J]]|$1|M|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]|$1|N|3|O|5|6|7|1G|8|@$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]]|D|@]|E|$]]|$1|P|3|Q|5|H|7|1L|8|@]|D|@]|E|$I|J]]|$1|R|3|S|5|6|7|1M|8|@$9|1N|A|1O|B|C]|$9|1P|A|1Q|B|C]|$9|1R|A|1S|B|C]|$9|1T|A|1U|B|C]|$9|1V|A|1W|B|C]]|D|@]|E|$]]|$1|T|3|U|5|6|7|1X|8|@$9|1Y|A|1Z|B|C]]|D|@]|E|$]]|$1|V|3|W|5|H|7|20|8|@]|D|@]|E|$I|J]]|$1|X|3|Y|5|H|7|21|8|@]|D|@]|E|$I|J]]|$1|Z|3|-4|5|6|7|22|8|@]|D|@]|E|$]]]|10|$11|$5|12|13|14|E|$15|16]]]]

How about using <a href="https://pandas.pydata.org/docs/reference/api/pandas.Series.map.html#pandas.Series.map" rel="nofollow noreferrer"><code>Series.map</code></a>:
<pre class="lang-py prettyprint-override"><code>import pandas as pd

df = pd.DataFrame({'x': ['good', 'bad', 'bad', 'good']})
print(df)
df['x'] = df.x.map({'good': 'bad', 'bad': 'good'})
print(df)
</code></pre>
<pre><code> x x
0 good 0 bad
1 bad → 1 good
2 bad 2 good
3 good 3 bad
</code></pre>
<hr />
For the edited part of the question, you can combine <code>Series.map</code> and <code>Series.where</code>:
<pre class="lang-py prettyprint-override"><code>import pandas as pd

df = pd.DataFrame({'x': ['good', 'bad', 'bad', 'good'], 'y': [100, 0, 100, 0]})
print(df)
df['x'] = df.x.map({'good': 'bad', 'bad': 'good'}).where(df.y &gt;= 100, df.x)
print(df)
</code></pre>
The idea is to swap the values globally (using <code>map</code>) and then revert back (using <code>where</code>) to the original <code>x</code> where the condition <code>df.y &gt;= 100</code> is <code>False</code>.
Alternatively, you can use <code>DataFrame.apply</code>:
<pre class="lang-py prettyprint-override"><code>swap = {'good': 'bad', 'bad': 'good'}
df['x'] = df.apply(lambda row: swap[row.x] if row.y &gt;= 100 else row.x, axis=1)
</code></pre>
<pre><code> x y x y
0 good 100 0 bad 100
1 bad 0 → 1 bad 0
2 bad 100 2 good 100
3 good 0 3 good 0
</code></pre>

blocks|key|1003240|text|您可以只使用pandas函数replace，它可以进行类似于dict的替换。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1003241|df.replace({'YourColumnName':+{'good':+'bad',++
+++++++++++++++++++++++++++++++'bad':+'good'}})|code-block|syntax|javascript|1003242|entityMap^0|6|6|E|7|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

You can just use the <code>pandas</code> function <code>replace</code> that can make dict-like replacements.
<pre><code>df.replace({'YourColumnName': {'good': 'bad', 
 'bad': 'good'}})
</code></pre>

blocks|key|3109813|text|你可以用replace来实现.|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3109814|df+=+pd.DataFrame({'A':+["good",+"good",+"bad",+"bad",+"good"],
+++++++++++++++++++'B':+[5,+6,+7,+8,+9],
+++++++++++++++++++'C':+['a',+'b',+'c',+'d',+'e'],
+++++++++++++++++++'D':+["good",+"good",+"bad",+"bad",+"good"],})


print(df.replace({'A':+{"good":+"bad",+"bad":+"good"}}))|code-block|syntax|javascript|3109815|这将仅限于列A。如果希望使用所有列，则可以在“替换”中解压字典。只需通过{"good":+"bad",+"bad":+"good"}|3109816|entityMap^0|4|7|0|0|6|1|10|U|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]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

You can use <code>replace</code> to achieve this...
<pre><code>df = pd.DataFrame({'A': [&quot;good&quot;, &quot;good&quot;, &quot;bad&quot;, &quot;bad&quot;, &quot;good&quot;],
 'B': [5, 6, 7, 8, 9],
 'C': ['a', 'b', 'c', 'd', 'e'],
 'D': [&quot;good&quot;, &quot;good&quot;, &quot;bad&quot;, &quot;bad&quot;, &quot;good&quot;],})


print(df.replace({'A': {&quot;good&quot;: &quot;bad&quot;, &quot;bad&quot;: &quot;good&quot;}}))
</code></pre>
This will limit to only column <code>A</code>. If you want all columns, you can unnest the dictionary in replace. and just pass <code>{&quot;good&quot;: &quot;bad&quot;, &quot;bad&quot;: &quot;good&quot;}</code>

blocks|key|1003291|text|使用np.where和@J1-Lee的数据帧：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1003292|import+numpy+as+np

df+=+pd.DataFrame({'x':+['good',+'bad',+'bad',+'good']})
df['y']+=+np.where(df['x'].eq('good'),'bad','good')|code-block|syntax|javascript|1003293|1003294|++++++x+++++y
0++good+++bad
1+++bad++good
2+++bad++good
3++good+++bad|1003295|entityMap^0|2|8|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|S|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|T|8|@]|D|@]|E|$]]|$1|L|3|M|5|H|7|U|8|@]|D|@]|E|$I|J]]|$1|N|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|O|$]]

Using <code>np.where</code> and @j1-lee's dataframe:
<pre><code>import numpy as np

df = pd.DataFrame({'x': ['good', 'bad', 'bad', 'good']})
df['y'] = np.where(df['x'].eq('good'),'bad','good')
</code></pre>
<hr />
<pre><code> x y
0 good bad
1 bad good
2 bad good
3 good bad
</code></pre>

blocks|key|1003337|text|这里有一个技巧，它可以像预期的那样对两个唯一值起作用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1003338|df['x'].replace({k:+v+for+k,+v+in+zip(df['x'].unique(),+df['x'].unique()[::-1])})|code-block|syntax|javascript|1003339|输出：|1003340|>>>+df
++++++x
0++good
1+++bad
2+++bad
3++good

>>>+df['x']+=+df['x'].replace({k:+v+for+k,+v+in+zip(df['x'].unique(),+df['x'].unique()[::-1])})
>>>+df
0+++++bad
1++++good
2++++good
3+++++bad|1003341|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's a trick that will work as expected for two unique values:
<pre class="lang-py prettyprint-override"><code>df['x'].replace({k: v for k, v in zip(df['x'].unique(), df['x'].unique()[::-1])})
</code></pre>
Output:
<pre><code>&gt;&gt;&gt; df
 x
0 good
1 bad
2 bad
3 good

&gt;&gt;&gt; df['x'] = df['x'].replace({k: v for k, v in zip(df['x'].unique(), df['x'].unique()[::-1])})
&gt;&gt;&gt; df
0 bad
1 good
2 good
3 bad
</code></pre>

I have dataframe with column which has 2 values: good and bad.
I want to know if there is a way that I can swap them. For example whenever value is good should be changed to bad or vice versa.
Is there a way to swap the values?
Edit: I want to swap values for certain indices after for loop.
Example:
<pre><code> for i in np.arange(0,len(df)-1):
 if(df.loc[i,'othercolumnname']&gt;=100:
 # swap the values here
</code></pre>

Pandas df swap column value

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个列的dataframe，它有两个值:好的和坏的。我想知道是否有办法可以交换它们。例如，无论何时值是好的，都应该改为坏的，反之亦然。有办法交换值吗？编辑:我想在for循环之后为某些索引交换值。示例： for i in np.arange(0,len(df)-1):     if(df.loc[i,'otherco...

问熊猫df交换列值
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问熊猫df交换列值EN