我有一个看起来像这样的矩阵：
M = [[1, 200],
[1.8, 100],
[2, 500],
[2.5, 300],
[3, 400],
[3.5, 200],
[5, 200],
[8, 100]]
我想按bin大小（应用于左侧列）对行进行分组，例如对于bin大小2（第一个bin的值是0-2，第二个bin的值是2-4，第三个bin的值是4-6，依此类推）：
[[1, 200],
[1.8, 100],
----
[2, 500],
[2.5, 300],
[3, 400],
[3.5, 200],
----
[5, 200],
----
[8, 100]]
然后输出一个新矩阵，其中包含每组右列的总和：
[200+100, 500+300+400+200, 200, 100]
基于bin_size边界求和每个值的有效方法是什么？
答案 0 :(得分：4)
apply plugin: "com.android.application"
apply plugin: 'kotlin-android'
apply plugin: "spoon"
apply plugin: "dexguard"
apply plugin: "io.fabric"
apply plugin: "jacoco"
apply plugin: 'realm-android'
apply plugin: 'kotlin-kapt'
apply plugin: 'kotlin-android-extensions'
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// versionName generated by the Jenkinsfile
fullVersion = project.realVersionName
} else {
// Fallback for local builds
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def versionCodeNr = Integer.parseInt(projectVersionCode)
androidExtensions {
experimental = true
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android {
compileSdkVersion 28
buildToolsVersion '28.0.3'
defaultConfig {
minSdkVersion 21
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versionName fullVersion
applicationId projectApplicationId
renderscriptTargetApi 28
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testInstrumentationRunner "android.support.test.runner.AndroidJUnitRunner"
ndk {
abiFilters "armeabi-v7a", "x86", "mips"
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buildTypes {
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report.initWith(buildTypes.debug)
report {
testCoverageEnabled true
matchingFallbacks = ['debug']
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client {
debuggable false
signingConfig signingConfigs.debug
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...
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targetCompatibility 1.8
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ext {
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...
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implementation "com.android.support:design:$supportLibVersion"
implementation "com.android.support:appcompat-v7:$supportLibVersion"
implementation "com.android.support:gridlayout-v7:$supportLibVersion"
implementation "com.android.support.constraint:constraint-layout:1.1.3"
implementation "com.google.firebase:firebase-core:16.0.1"
implementation "com.google.firebase:firebase-messaging:17.1.0"
//noinspection GradleDependency
implementation "com.google.code.gson:gson:2.8.5"
// Glide
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apply plugin: "com.google.gms.google-services"
：制作一个pandas，然后使用整数除法来定义垃圾箱：
DataFrame使用import pandas as pd
df = pd.DataFrame(M)
df.groupby(df[0]//2)[1].sum()
#0
#0.0 300
#1.0 1400
#2.0 200
#4.0 100
#Name: 1, dtype: int64
获得所需的输出：
.tolist()df.groupby(df[0]//2)[1].sum().tolist()
#[300, 1400, 200, 100]
numpy.bincount答案 1 :(得分：2)
您可以在这里使用np.digitize和scipy.sparse.csr_matrix：
bins = [2, 4, 6, 8, 10]
b = np.digitize(M[:, 0], bins)
v = M[:, 1]
现在使用矢量groupby和csr_matrix
from scipy import sparse
sparse.csr_matrix(
(v, b, np.arange(v.shape[0]+1)), (v.shape[0], b.max()+1)
).sum(0)
matrix([[ 300., 1400., 200., 0., 100.]])