blocks|key|2257198|text|像这样的东西会起作用的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2257199|df.loc[df['Fruit']+==+df['Fruit_Copy'],+'Match']+=+'Yes'|code-block|syntax|javascript|2257200|使用numpy.where:|offset|length|2257201|df['Match']+=+np.where(df['Fruit']+==+df['Fruit_Copy'],+'Yes',+'No')|2257202|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.scipy.org/doc/numpy/reference/generated/numpy.where.html^0|0|0|2|C|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@$I|X|J|Y|1|Z]]|A|$]]|$1|K|3|L|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

Something like this would work.

<pre><code>df.loc[df['Fruit'] == df['Fruit_Copy'], 'Match'] = 'Yes'
</code></pre>

Using <a href="https://docs.scipy.org/doc/numpy/reference/generated/numpy.where.html" rel="nofollow noreferrer">numpy.where:</a>

<pre><code>df['Match'] = np.where(df['Fruit'] == df['Fruit_Copy'], 'Yes', 'No')
</code></pre>

blocks|key|216489|text|您可以尝试如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|216490|import+pandas+as+pd
import+numpy+as+np

fruits+=+pd.DataFrame({'Fruit':['Apple',+'Orange',+'Apple',+'Apple',+'Banana',+'Banana',+'Apple',+'Apple'],+'Cost':[0.5,0.3,0.5,0.5,0.25,0.25,0.5,0.5],+'Quantity':[6,2,8,7,8,7,6,3],+'Fruit_Copy':['Watermelon',+'Orange',+'Apple',+'Apple',+'Banana',+'Banana',+'Apple',+'Apple']})
fruits['Match']+=+np.where(fruits['Fruit']+==+fruits['Fruit_Copy'],+1,+0)
fruits

++++Fruit++Cost++Quantity++Fruit_Copy++Match
0+++Apple++0.50+++++++++6++Watermelon++++++0
1++Orange++0.30+++++++++2++++++Orange++++++1
2+++Apple++0.50+++++++++8+++++++Apple++++++1
3+++Apple++0.50+++++++++7+++++++Apple++++++1
4++Banana++0.25+++++++++8++++++Banana++++++1
5++Banana++0.25+++++++++7++++++Banana++++++1
6+++Apple++0.50+++++++++6+++++++Apple++++++1
7+++Apple++0.50+++++++++3+++++++Apple++++++1|code-block|syntax|javascript|216491|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|$]]

You could try something like this:

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

fruits = pd.DataFrame({'Fruit':['Apple', 'Orange', 'Apple', 'Apple', 'Banana', 'Banana', 'Apple', 'Apple'], 'Cost':[0.5,0.3,0.5,0.5,0.25,0.25,0.5,0.5], 'Quantity':[6,2,8,7,8,7,6,3], 'Fruit_Copy':['Watermelon', 'Orange', 'Apple', 'Apple', 'Banana', 'Banana', 'Apple', 'Apple']})
fruits['Match'] = np.where(fruits['Fruit'] == fruits['Fruit_Copy'], 1, 0)
fruits

 Fruit Cost Quantity Fruit_Copy Match
0 Apple 0.50 6 Watermelon 0
1 Orange 0.30 2 Orange 1
2 Apple 0.50 8 Apple 1
3 Apple 0.50 7 Apple 1
4 Banana 0.25 8 Banana 1
5 Banana 0.25 7 Banana 1
6 Apple 0.50 6 Apple 1
7 Apple 0.50 3 Apple 1
</code></pre>

blocks|key|216485|text|假设你的数据帧是“水果”。然后你可以使用熊猫系列等于函数pd.Series.eq+as，|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|216486|fruits['Match']+=+pd.Series.eq(fruits['Fruit'],fruits['Fruit_Copy'])|code-block|syntax|javascript|216487|entityMap|0|LINK|mutability|MUTABLE|url|https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.eq.html#pandas.Series.eq^0|S|C|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]]]]

Lets say your dataframe is 'fruits'. Then you can make use of the Pandas Series Equals function <a href="https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.eq.html#pandas.Series.eq" rel="nofollow noreferrer">pd.Series.eq</a> as,

<pre><code>fruits['Match'] = pd.Series.eq(fruits['Fruit'],fruits['Fruit_Copy'])
</code></pre>

I have a dataframe that looks like this:

<pre><code>Fruit Cost Quantity Fruit_Copy
Apple 0.5 6 Watermelon
Orange 0.3 2 Orange
Apple 0.5 8 Apple
Apple 0.5 7 Apple
Banana 0.25 8 Banana
Banana 0.25 7 Banana
Apple 0.5 6 Apple
Apple 0.5 3 Apple
</code></pre>

I want to write a snippet that, in pandas, compares Fruit and Fruit_Copy and outputs a new column "Match" that indicates if the values in Fruit = Fruit_Copy.

Thanks in advance!

Compare values in two different pandas columns

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。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我有一个数据帧，看起来像这样：Fruit   Cost    Quantity    Fruit_CopyApple   0.5 6   WatermelonOrange  0.3 2   OrangeApple   0.5 8   AppleApple   0.5 7   AppleBanana  0.25    8...

问比较两个不同pandas列中的值
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问比较两个不同pandas列中的值EN