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单片机(MCU)如何才能不死机之对齐访问(Aligned Access)

这是 ARM Cortex M0 体系决定的,它只支持对齐访问 ( Aligned Access )。

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乐固-签名APK-会出来一个aligned的apk包,在签名的时候你们还做了什么操作吗?

乐固-签名APK-会出来一个aligned的apk包,在签名的时候你们还做了什么操作吗?

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    MOT:Metrics HOTA

    } \text {MOTA} = 1 - \frac{|\text {FN}| + |\text {FP}| + |\text {IDSW}|}{|\text {gtDet}|} \end{aligned } \text {MOTP} = \frac{1}{|\text {TP}|}\sum _{\text {TP}}{ \mathcal {S}} \end{aligned} } \begin{aligned} \text {MODA}&= 1 - \frac{|\text {FN}| + |\text {FP}|}{|\text {gtDet}|}&\\&= \frac{| } \end{aligned} 可以发现,如果检测的Precision小于等于0.5的话,MODA就会为0,甚至出现负值,而检测的Recall小于等于0.5 }&\text {DetA}_\alpha = \frac{|\text {TP}|}{|\text {TP}| + |\text {FN}| + |\text {FP}|}&\end{aligned}

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    MKL sparse QR solver for least square

    , int *cols_aligned, float *vals_aligned, int num_rows, int num_cols, const size_t num_non_zeros) { = std_vec_2_aligned_array<int>(i_idx, mem_align); int *cols_aligned = std_vec_2_aligned_array<int (); sparse_matrix_t coo_mtx = create_coo_mtx(rows_aligned, cols_aligned, vals_aligned, num_rows, ); free(cols_aligned); free(vals_aligned); free(b_aligned); free(x); #else _aligned_free (rows_aligned); _aligned_free(cols_aligned); _aligned_free(vals_aligned); _aligned_free(b_aligned

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    金黄葡萄球菌RNA-seq数据分析

    concordantly 0 times 13590 (0.10%) aligned concordantly exactly 1 time 7230 (0.05%) aligned 1695827 (27.69%) aligned exactly 1 time 4432 (0.07%) aligned >1 times 86.16% overall 1622752 (27.93%) aligned exactly 1 time 7228 (0.12%) aligned >1 times 85.29% overall 2110960 (27.05%) aligned exactly 1 time 20696 (0.27%) aligned >1 times 84.48% overall 1838519 (26.75%) aligned exactly 1 time 8174 (0.12%) aligned >1 times 85.32% overall

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    MOT:A Higher Order Metric for Evaluating Multi-object Tracking

    } \text {MOTA} = 1 - \frac{|\text {FN}| + |\text {FP}| + |\text {IDSW}|}{|\text {gtDet}|} \end{aligned } \begin{aligned} \text {MODA}&= 1 - \frac{|\text {FN}| + |\text {FP}|}{|\text {gtDet}|}&\\&= \frac{| } \end{aligned} MODA​=1−∣gtDet∣∣FN∣+∣FP∣​=∣TP∣+∣FN∣∣TP∣−∣FP∣​=DetRe⋅(2−DetPr1​)​​​ }&\text {DetA}_\alpha = \frac{|\text {TP}|}{|\text {TP}| + |\text {FN}| + |\text {FP}|}&\end{aligned} }&\begin{aligned} \text {HOTA}_\alpha&\quad = \sqrt{\frac{\sum _{c \in \{\text {TP}\}} \mathcal {A}(c

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    傅里叶级数

    \end{aligned} \tag{2} 这里面的an,bn系数怎么求呢? +c \end{aligned} \tag{4} 有了上面这些基本公式,先来证明几个恒等式: \begin{aligned} \int_{0}^{2\pi}sin(mt)dt &=0 , (m \ne 0) \end{aligned} \tag{a} \begin{aligned} \int_{0}^{2\pi}cos(mt)dt &=0,(m \ne 0) \\ \end {aligned} \tag{b} \begin{aligned} \int_{0}^{2\pi}sin(mt) \cdot cos(nt)dt &=0,(m \ne n , m \ne -n ) \\ \end{aligned} \tag{c} \begin{aligned} \int_{0}^{2\pi}cos(mt) \cdot cos(nt)dt &=0,(m \ne n

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    是人是鼠,你心里没有数?GPL21827之谜

    比如GPL21827这个平台: 60898 reads; of these: 60898 (100.00%) were unpaired; of these: 59099 (97.05%) aligned 0 times 1753 (2.88%) aligned exactly 1 time 46 (0.08%) aligned >1 times 2.95% overall alignment 0 times 811 (0.48%) aligned exactly 1 time 149 (0.09%) aligned >1 times 0.56% overall alignment acc=GPL8180 380 reads; of these: 380 (100.00%) were unpaired; of these: 215 (56.58%) aligned 0 times 138 (36.32%) aligned exactly 1 time 27 (7.11%) aligned >1 times 43.42% overall alignment

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    sar - Linux 系统监控利器

    ((aligned (8))); unsigned long InEchoReps __attribute__ ((aligned (8))); unsigned long OutEchos __ attribute__ ((aligned (8))); unsigned long OutEchoReps __attribute__ ((aligned (8))); unsigned long InTimestamps __attribute__ ((aligned (8))); unsigned long InTimestampReps __attribute__ ((aligned ( __attribute__ ((aligned (8))); unsigned long OutAddrMasks __attribute__ ((aligned (8))); unsigned __attribute__ ((aligned (8))); unsigned long OutTimeExcds __attribute__ ((aligned (8))); unsigned

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    大肠杆菌全基因组重测序变异检测小实例(侧重变异过滤)

    学着在shell下写循环 cd output_results #SAM装换为BAM samtools view -S -b -o sim_1_aligned.bam sim_1_aligned.sam samtools view -S -b -o sim_2_aligned.bam sim_2_aligned.sam samtools view -S -b -o sim_3_aligned.bam sim_3_aligned.sam #排序 samtools sort sim_1_aligned.bam -o sim_1_aligned.sorted.bam samtools sort sim_ 2_aligned.bam -o sim_2_aligned.sorted.bam samtools sort sim_3_aligned.bam -o sim_3_aligned.sorted.bam /Reference_genome/ecoli.fa sim_1_aligned.sorted.bam sim_2_aligned.sorted.bam sim_3_aligned.sorted.bam

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    【STL源码拆解】基于源码分析forward_lsit容器实现(详细!)

    _M_ptr(); } }; 该类定义了一个成员变量_M_storage,这个成员变量类型是__gnu_cxx::__aligned_buffer<_Tp>,同样是一个类模板,看看类模板__aligned_buffer 的类声明,如下: template<typename _Tp> struct __aligned_buffer : std::aligned_storage<sizeof(_Tp), std ::alignment_of<_Tp>::value> { typename std::aligned_storage<sizeof(_Tp), std::alignment_of __(typename __aligned_storage_msa<_Len>::__type)> struct aligned_storage { union type }; 基于以上代码,可以看出类型aligned_storage::type其实是一个联合体类型,该联合体的第一个字段很明朗,就是以模板类型_Tp的长度定义了一个无符号字符数组,但第二个字段__align

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    samtools小实例(未完成)

    sorted.bam samtools index eg2.sorted.bam #过滤没有比对到参考基因组的reads samtools view -F 4 eg2.sorted.bam > eg2.aligned.sam #根据fasta文件将header添加到sam文件中 samtools view -T reference_genome.fasta -h eg2.aligned.sam > eg2.aligned.header.sam #SMA转BAM samtools view -b -S -o eg2.aligned.header.bam eg2.aligned.header.sam #BAM转换fastq bam2fastq --aligned -o aligned.fastq eg2.aligned.header.sam samtools 常用操作 #没有比对到参考基因组上reads的数量 samtools view - c -f -4 aligned.sorted.bam #sam转bam samtools view -b -S -o A.bam A.sam #排序 samtools sort A.bam -o A.sorted.bam

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    序列比对(15)EM算法以及Baum-Welch算法的推导

    theta^{t+1}) - \mathrm{log} \, P(x|\theta^t) \geq 0 \tag{3}$$ 为了得到$\theta^{t+1}$,我们首先变换得到: $$\begin{aligned \mathrm{log} \, P(x,y|\theta)\\ & -\sum_y P(y|x,\theta^t)\mathrm{log} \, P(y|x,\theta) \tag{4} \end{aligned }$$ >公式(4)可以这样推导得到: 将公式(2)等式两边乘上$P(y|x,\theta^t)$,再对所有的$y$求和,得到: $$\begin{aligned} \displaystyle \sum_y }$$ 很容易看出等式左边: $$\begin{aligned} \displaystyle \sum_y P(y|x,\theta^t)\mathrm{log} \, P(x|\theta) & = }$$ >公式(6)的推导也很容易: $$\begin{aligned} &\mathrm{log} \, P(x|\theta) - \mathrm{log} \, P(x|\theta^t)\\

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    ChIP-seq详细分析流程

    (39.51%) aligned 0 times 7766495 (39.45%) aligned exactly 1 time 4142095 (21.04%) aligned >1 (12.48%) aligned 0 times 10914615 (52.70%) aligned exactly 1 time 7210683 (34.82%) aligned > (32.99%) aligned 0 times 9455003 (43.87%) aligned exactly 1 time 4987072 (23.14%) aligned >1 (15.83%) aligned 0 times 20702502 (51.92%) aligned exactly 1 time 12855241 (32.24%) aligned (42.20%) aligned 0 times 4663584 (35.09%) aligned exactly 1 time 3019049 (22.71%) aligned >1

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    Sweet Snippet 系列之 扩展欧几里得算法

    yyy 的递推公式,以方便我们编写代码: 首先是基础条件(b=0b = 0b=0 的情况) gcd(a,0)=aa∗1+0∗any=gcd(a,0)=a=>{x=1y=0 \begin{aligned } & gcd(a, 0) = a \\ & a * 1 + 0 * any = gcd(a, 0) = a => \\ \end{aligned} \\ \left\{ \begin{aligned} & x = 1 \\ & y = 0 \end{aligned} \right. ​gcd } \\ \left\{ \begin{aligned} & x = y' \\ & y = x' - \lfloor a / b \rfloor y' \end{aligned} \right. ​ax+by=gcd(a,b)bx′+(a%b)y′=gcd(b,a%b)∵gcd(a,b)=gcd(b,a%

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    矩阵分析复习题

    mathbb{F})中非零向量,若\mathbb{F}中数k_1与k_2不相等,试证:k_1\alpha\neq k_2\alpha 证:假设k_1\alpha =k_2\alpha,则 $$ \begin{aligned } k_1\alpha-k_2\alpha=0 \\ \Rightarrow (k_1-k_2)\alpha=0 \end{aligned} $$ 因为\alpha \neq 0,所以k_1-k_2=0 } $$ 因为 $$ \begin{aligned} A^HA &= (BC)^H(BC)\\ &=C^HB^HBC\\ &=(B^HB)(C^HC)\quad(因为B^HB是一个数) \end{aligned } $$ 所以 $$ \begin{aligned} tr(A^HA) &= B^HB·tr(C^HC)\\ &=B^HB ·tr(CC^H)\\ &=B^HB·CC^H \end{aligned} $ ,&\quad {0≤m≤n}\\0, &\quad {others}\end{cases} \end{aligned} $$

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    转录组上游分析错误(报错大赏)

    aligned concordantly >1 times ---- 1295673 pairs aligned concordantly 0 times; of these: 47378 (3.66%) aligned discordantly 1 time ---- 1248295 pairs aligned 0 times concordantly or 0 times 900163 (36.06%) aligned exactly 1 time 141180 (5.65%) aligned >1 times 96.90% aligned concordantly >1 times ---- 2727958 pairs aligned concordantly 0 times; of these: 32392 (1.19%) aligned discordantly 1 time ---- 2695566 pairs aligned 0 times concordantly or

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    数学小记之常用数值

    100732835 本文简单列举了一些常用数值,熟记这些数值可以方便我们进行数学运算 π\piπ 相关的数值想必大部分朋友都比较熟悉了: π≈3.14π2≈1.572π≈6.28 \begin{aligned amp; \pi \approx 3.14 \\ & \frac{\pi}{2} \approx 1.57 \\ & 2\pi \approx 6.28 \end{aligned 2.718 e≈2.718 一些常用的根号值: 1=12≈1.4143≈1.7324=25≈2.2366≈2.449497≈2.645758=22≈2.8289=310≈3.162 \begin{aligned } = 2\sqrt{2} \approx 2.828 \\ & \sqrt{9} = 3 \\ & \sqrt{10} \approx 3.162 \end{aligned } & log_{10}2 \approx 0.3010 \\ & log_{10}3 \approx 0.4771 \end{aligned} ​log10​2

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    深度学习中的激活函数总结

    g(z) =\left\{ \begin{aligned } z , if :z>0 \\ z , if : z<0 \end{aligned} \right. } 1 , if :z>0 \\ 0 , if : z<0 \end{aligned} \right. } z , if :z>0 \\ az , if : z<0 \end{aligned} \right. } 1 , if :z>0 \\ a , if : z<0 \end{aligned} \right.

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    RNA-seq(5):序列比对:Hisat2

    concordantly 0 times 24733251 (85.71%) aligned concordantly exactly 1 time 2284771 (7.92%) aligned 1221304 (34.94%) aligned exactly 1 time 239649 (6.86%) aligned >1 times 96.47% overall aligned concordantly >1 times ---- 2722534 pairs aligned concordantly 0 times; of these: 156866 (5.76%) aligned discordantly 1 time ---- 2565668 pairs aligned 0 times concordantly or 0 times 1334397 (26.00%) aligned exactly 1 time 520406 (10.14%) aligned >1 times 94.62%

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