我想将列表拆分为偶数大小的块，其中n-max作为参数。如果n-max大于列表长度，则将列表拆分为最大的重叠块。
示例：
l = [1,2,3,4,5,6,7,8,9]
n-max = 4
for k, v in enumerate (l):
if k < len(l):
a = l[k:k+n-max]
if len(a) == n-max:
yield a
>>> [[1,2,3,4],[2,3,4,5],[3,4,5,6,. [4,5,6,7],[5,6,7,8],[6,7,8,9]]
当l的长度小于n-max时。我希望它返回最大重叠块，
l = [1,2,3]
>>> [[1,2],[2,3]]
最大重叠块大小为2.如果列表无法在n-max处进行块化，我需要使用函数自动将减少n-max块大小设置为列表可能的重叠块大小。
答案 0 :(得分：3)
作为一种pythonic方式，您可以使用列表推导来分割列表，方法是选择<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
<modelVersion>4.0.0</modelVersion>
<groupId>com.dario.apachepoi</groupId>
<artifactId>apachepoi</artifactId>
<version>0.0.1-SNAPSHOT</version>
<packaging>jar</packaging>
<properties>
<project.build.sourceEncoding>UTF-8</project.build.sourceEncoding>
<maven.compiler.source>1.8</maven.compiler.source>
<maven.compiler.target>1.8</maven.compiler.target>
</properties>
<dependencies>
<!-- https://mvnrepository.com/artifact/org.apache.poi/poi -->
<dependency>
<groupId>org.apache.poi</groupId>
<artifactId>poi</artifactId>
<version>3.15</version>
</dependency>
<!-- https://mvnrepository.com/artifact/org.apache.poi/poi-examples -->
<dependency>
<groupId>org.apache.poi</groupId>
<artifactId>poi-examples</artifactId>
<version>3.15</version>
</dependency>
<!-- https://mvnrepository.com/artifact/org.apache.poi/poi-excelant -->
<dependency>
<groupId>org.apache.poi</groupId>
<artifactId>poi-excelant</artifactId>
<version>3.15</version>
</dependency>
<!-- https://mvnrepository.com/artifact/org.apache.poi/poi-ooxml -->
<dependency>
<groupId>org.apache.poi</groupId>
<artifactId>poi-ooxml</artifactId>
<version>3.15</version>
</dependency>
<!-- https://mvnrepository.com/artifact/org.apache.poi/poi-ooxml-schemas -->
<dependency>
<groupId>org.apache.poi</groupId>
<artifactId>poi-ooxml-schemas</artifactId>
<version>3.15</version>
</dependency>
<!-- https://mvnrepository.com/artifact/org.apache.poi/poi-scratchpad -->
<dependency>
<groupId>org.apache.poi</groupId>
<artifactId>poi-scratchpad</artifactId>
<version>3.15</version>
</dependency>
<!-- https://mvnrepository.com/artifact/org.apache.xmlbeans/xmlbeans -->
<dependency>
<groupId>org.apache.xmlbeans</groupId>
<artifactId>xmlbeans</artifactId>
<version>2.6.0</version>
</dependency>
<!-- https://mvnrepository.com/artifact/com.github.virtuald/curvesapi -->
<dependency>
<groupId>com.github.virtuald</groupId>
<artifactId>curvesapi</artifactId>
<version>1.04</version>
</dependency>
<!-- https://mvnrepository.com/artifact/org.apache.commons/commons-lang3 -->
<dependency>
<groupId>org.apache.commons</groupId>
<artifactId>commons-lang3</artifactId>
<version>3.0</version>
</dependency>
</dependencies>
<build>
<finalName>ReadDocFile</finalName>
<plugins>
<!-- download source code in Eclipse, best practice -->
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-eclipse-plugin</artifactId>
<version>2.10</version>
<configuration>
<downloadSources>true</downloadSources>
<downloadJavadocs>false</downloadJavadocs>
</configuration>
</plugin>
<!-- Set a compiler level -->
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-compiler-plugin</artifactId>
<version>3.3</version>
<configuration>
<source>${jdk.version}</source>
<target>${jdk.version}</target>
</configuration>
</plugin>
<!-- Maven Shade Plugin -->
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-shade-plugin</artifactId>
<version>2.4.3</version>
<executions>
<!-- Run shade goal on package phase -->
<execution>
<phase>package</phase>
<goals>
<goal>shade</goal>
</goals>
<configuration>
<transformers>
<!-- add Main-Class to manifest file -->
<transformer implementation="org.apache.maven.plugins.shade.resource.ManifestResourceTransformer">
<mainClass>com.dario.apachepoi.PrintSomething</mainClass>
</transformer>
</transformers>
</configuration>
</execution>
</executions>
</plugin>
</plugins>
</build>
</project>
和n的最小值作为块大小。
len(your_list)-1演示：
In [39]: def chunk(n, lst):
n = min(n, len(lst)-1)
return [lst[i:i+n] for i in range(len(lst) - n+1)]