Problem_Set_1ques
python中set的用法和作用
python中set的用法和作用
在Python中，set是一种无序且元素不重复的数据结构。
它可以用来存储一组唯一的元素，类似于数学中的集合概念。
在Python 中，set使用花括号{}来表示，其中的元素之间用逗号分隔。
set的主要作用之一是用于去除列表中重复的元素。
通过将一个列表转换为set，然后再转换回列表，就可以轻松去除重复的元素。
这在需要对数据进行去重操作时非常有用。
另外，set还可以用于数学运算，如并集、交集、差集等。
通过使用set提供的方法，可以方便地进行这些集合运算。
例如，可以使用union()方法求两个集合的并集，intersection()方法求两个集合的交集，difference()方法求两个集合的差集等。
此外，set还可以用于快速判断一个元素是否属于某个集合，因为set内部使用了哈希表来存储元素，所以判断元素是否存在的速度非常快。
需要注意的是，由于set是无序的，所以不能像列表一样通过索引来访问元素。
另外，由于set内部的元素不重复，所以不能向
set中添加重复的元素。
总之，set在Python中具有去除重复元素、集合运算和快速判断元素是否存在等作用，是一个非常实用的数据结构。
Problem_Set_9ques
Harvard University Economics1123Department of Economics Spring2006Problem Set9Perhaps they need bigger subsidies.Due:Thursday4May.Gas prices have hit$3,many analysts consider this a sign or portent or both.Oil executives argue that oil prices are a competitive market,and that the recent record profits(Chevron reported a$14.1billion profit for last year)are just the up part in the usual ups and downs of the free market.The federal government stepped in last year to help out,passing an energy bill that gave an estimated$2.6billion in tax breaks to the companies.Consumers argue that something else is going on,consumer groups complain of’gouging’and the federal government has today eased pollution laws for oil companies and announced hearings into oil pricing.We have data on West Texas Intermediate(WTI)crude oil(Cushing OK spot prices)monthly from June1986until March of this year(the latest available end month number),denoted poil in the dataset.The data is dollars per barrel.We also have data on the US gulf coast regular spot price,i.e.the cost of gas(at the refinery)in cents per gallon.This is denoted pgas.In this problem set we are going to analyze the data and produce a forecasting model using the data for each of the variables.We also consider issues of causality.First,construct logs of the data,and thenfirst differences.The data are nonstationary,so this seems like a reasonable approach.Question1.Run AR(1)regressions for both the variables infirst differences.Construct the AR(1)forecasts of the April WTI and gas prices and report them.Question2.Run VAR regressions with a single lag for each of the variables.Test(a)if gas prices Granger cause Oil Prices(b)if oil prices Granger cause gas prices.(c)do the results make much sense?Comment with regards to causality,do not write a long essay.Question3.Gas prices and oil prices follow each other closely,hence one might expect that they are coin-tegrated,i.e.the difference between them is a stationary variable and hence can be used in the predictive regression.Construct this variable and test whether or not it has a unit root(use a constant but not a trend in the test).Report yourfindings as to whether or not this variable has a unit root or not.Justify your lag length choice.Question4.Run the same regressions as in Question2but now add the error correction term(remember to lag it one period).Construct forecasts of the two prices for April and report them.Do you think they are likely to be more reasonable?At the EIA home page at the DOE"This week in Petroleum"should give the updated WTI price to compare with the forecasts.There is an update due Wed26April(tomorrow at the time of writing).You can check your forecasts against this!Hints.In stata,for a time series variable oil,you can tell state to lag it j times by referring to’lj.oil’. This can be used with the generate command and the regression command,you just refer to it as though you had created the variable.Stata is careful to drop initial observations so you do not have to worry about that.I have put the EXCELfile oil.txt on the web in case you prefer to use this to generate the forecasts.。
MATLAB中常见问题解决方案大全
MATLAB中常见问题解决方案大全引言：MATLAB是一种功能强大的数学计算软件，广泛应用于科学研究、工程设计和数据分析等领域。
然而，在使用MATLAB的过程中，我们经常会遇到一些问题和困惑。
本文将总结一些常见的MATLAB问题，并提供相应的解决方案，帮助读者更好地理解和应用这个工具。
一、MATLAB的安装问题解决方案1. 问题描述：安装MATLAB时遇到许可证问题。
解决方案：首先，确保已经获取到了有效的许可证文件。
然后，运行安装程序并按照提示进行操作。
若仍出现问题，可以尝试禁用防火墙、关闭杀毒软件，并以管理员身份运行安装程序。
2. 问题描述：安装过程中出现错误代码。
解决方案：错误代码通常会提供问题的具体描述，可通过MATLAB官方网站或谷歌搜索相关错误代码进行查找。
MATLAB官方网站提供了相应的解决方案和技术支持。
二、MATLAB的基础问题解决方案1. 问题描述：如何导入和保存数据？解决方案：可以使用`load`函数导入数据，使用`save`函数保存数据。
另外，MATLAB还支持其他格式的数据导入和导出，如`csvread`和`csvwrite`用于CSV格式，`xlsread`和`xlswrite`用于Excel格式等。
2. 问题描述：如何修改MATLAB的默认设置？解决方案：可以通过修改MATLAB的配置文件来实现。
通过运行命令`edit('matlabrc.m')`可以打开该文件，并根据需要修改默认设置。
三、MATLAB的数据处理问题解决方案1. 问题描述：如何处理丢失数据？解决方案：可以使用MATLAB提供的插值函数来处理丢失数据，如`interp1`和`interp2`等。
这些函数可以根据已有数据的趋势，推断出丢失数据的可能取值，从而填补空缺。
2. 问题描述：如何处理异常值？解决方案：可以使用MATLAB中的统计函数来处理异常值，如`mean`和`median`等。
python编程中常见错误
python编程中常见错误python中常见错误最后，我想谈谈使⽤更多python函数（数据类型、函数、模块、类等）时可能遇到的问题。
由于篇幅有限，我们试图将其简化，特别是⼀些⾼级概念。
有关更多详细信息，请阅读学习python、第⼆版的“技巧”和“gotchas”章节。
打开⽂件的调⽤不使⽤模块搜索路径在python中调⽤open（）访问外部⽂件时，python不使⽤模块搜索路径来定位⽬标⽂件。
它将使⽤您提供的绝对路径，或者假定⽂件位于当前⼯作⽬录中。
模块搜索路径仅⽤于模块加载。
不同的类型对应不同的⽅法。
列表⽅法不能⽤于字符串，反之亦然。
通常，⽅法调⽤与数据类型相关，但内部函数通常在许多类型上可⽤。
例如，list的reverse⽅法只对list有⽤，但是len函数对任何有长度的对象都有效。
⽆法直接更改不可变数据类型请记住，不能直接更改不可变对象（例如元组、字符串）：t=1、2、3t[2]=4误差⽤切⽚、连接等构造⼀个新的对象，并根据需要为其分配原始变量的值。
因为python会⾃动回收⽆⽤的内存，所以并不像看起来那么浪费：t=t[：2]+（4，）没问题：t变成（1，2，4）使⽤简单for循环⽽不是while或range当您从左到右遍历有序对象的所有元素时，使⽤简单的for循环（例如，seq:中的x）⽐使⽤基于while或range的计数循环更容易编写，并且通常运⾏得更快。
除⾮需要，否则请尽量避免在for循环中使⽤range：让python为您解决标记问题。
在下⾯的⽰例中，这三个循环都很好，但第⼀个循环通常更好；在python中，简单性是最重要的。
S=伐⽊⼯⼈对于c in s:print c是最简单的对于范围内的i（len（s）：打印s[i]太多I=0太多当i<len（s）：打印s[i]；i+=1不要试图从更改对象的函数中获取结果。
直接更改操作，如⽅法列表。
追加（）并列出。
sort（）更改对象，但不返回更改的对象（它们不返回任何对象）；正确的⽅法是直接调⽤它们，⽽不是分配结果。
示教器操作手册最终版
MATLAB中常见错误及解决方法汇总
MATLAB中常见错误及解决方法汇总MATLAB是一种强大的数值计算和科学计算软件，被广泛应用于工程、科学和数学领域。
然而，在使用MATLAB时，我们常常会遇到一些错误和问题。
本文将汇总一些常见的MATLAB错误，并提供解决方法，帮助读者更好地处理和解决这些问题。
1. 向量维度不匹配错误这是在进行向量运算或矩阵操作时经常遇到的错误。
当出现该错误时，通常是因为参与运算的向量或矩阵的维度不匹配。
解决方法是检查参与运算的向量或矩阵的维度，确保其维度一致才能进行运算。
2. 数组索引越界错误当我们使用索引访问数组的元素时，如果指定的索引值超过了数组的大小范围，就会产生数组索引越界错误。
解决方法是检查索引值，并确保它们在数组大小范围内。
3. 未找到某个函数或变量的错误当我们尝试调用一个不存在的函数或访问一个未定义的变量时，就会产生未找到某个函数或变量的错误。
解决方法是检查函数或变量的名称是否正确拼写，并确保它们存在于当前工作空间或已添加到MATLAB的搜索路径中。
4. 内存不足错误大规模计算或处理复杂数据时，有时会出现内存不足的错误。
解决方法包括：- 减少数据的规模或精度；- 释放已使用的内存空间；- 使用更高配置的计算机或服务器。
5. 文件读写错误在进行文件读写操作时，常常会遇到文件读写错误。
解决方法包括：- 检查文件路径和名称是否正确；- 确保文件具有正确的读写权限；- 关闭已打开的文件或释放文件资源。
6. 函数参数个数不匹配错误在调用函数时，如果提供的参数个数与函数定义的参数个数不匹配，就会产生函数参数个数不匹配错误。
解决方法是检查函数的定义，并确保提供的参数个数和类型与定义一致。
7. 函数或脚本文件未结束错误在编写函数或脚本文件时，如果忘记在文件末尾添加"end"关键字，就会产生函数或脚本文件未结束错误。
解决方法是在文件的适当位置添加"end"关键字，以标识函数或脚本文件的结束。
系统生物学Ques-1.1 ans
7.36/7.91/20.390/20.490/6.802/6.874PROBLEM SET 1. Sequence search, global alignment, BLAST statistics (19 Points).Due: Thursday, February 20th at noonProblem 1. Sequence search (6 points)To better understand inborn disorders of metabolism, you isolate a strain of mice that becomes ill unless fed a diet lacking phenylalanine. You sequence the genome of this mouse and find several differences from wildtype including a change to a region that encodes a highly expressed 68 nucleotide RNA which has sequence5’-UGUACAUGAUGAAGUCAUAGCGAACGGAGAAGGGCCGGCUGAGGAA ACUGCACGUCACCCUCCUGAAA-3’in your strain and5’-UGUACAUGAUGAAAACAGUCUCCCUCUUCUGAAUCUCGCUGAGGAA ACUGCACGUCACCCUCCUGAAA-3’in wildtype mice.Search the sequence in your strain against the mouse genome and transcriptome using NCBI’s BLASTn: from the BLAST homepage, click on “nucleotide blast” (not “Mouse”) and use the “Mouse genomic + transcript” (G+T) Database, optimized for “Somewhat similar sequences”. By expanding the “Algorithm Parameters” box at the bottom, set the Match/Mismatch scores to +1/-3.(A) (1 pt.) How many statistically significant hits are there at an E-value of 0.05? In one sentence, what does an E-value of 0.05 mean? For transcript hits, what are the maximum reported scores, and are they raw scores or bit scores? (Click on the hyperlink to view individual hits.) To what parts of your RNA do these hits correspond, and what is the % match?There are two transcript and two genome hits at an E-value of 0.05. The E-value is the expected number of hits with score at least as high as the hit’s reported score when searching a query of length 68 nt against the Mouse G+T database. The maximum scores for the two transcript hits are 54 and 50.1 bits. The hit with score 54 bits corresponds to positions 38-68 of the query and has 97% identity to its match (matches 30 of 31 positions), while the hit with score 50.1 bits corresponds to positions 14-38 of the query and has 100% identity.(B) (1 pt.) Using the E-value and reported score from the result with the highest % identity match from part (A), calculate the approximate length of the Mouse (G+T) Database.Using the score S = 50.1 bits and E-value = 2 yields a mouse G+T Database length of ∗10E −value =mn 2!!mouse haploid genome assembly is about 2.7 billion base pairs, so after adding in tra !! along with m = 68nt in the formula n =3.55∗10!. Note that the nscriptsequences, the estimate from the formula is around what we would expect (various corrections to the simple formula are made for base content, repetitive regions, and other parameters for the reported BLAST values).(C) (1 pt.) Consider a query sequence Q of length L that matches perfectly to a sequence in the database, yielding a BLAST E-value E 1. How would the E-value change if only the first half of Q were searched against the database? In particular, would it stay the same, go up, go down, and how (linearly, exponentially, etc.)?Intuitively, decreasing the length of query (and therefore match) should make the match more likely simply by chance and therefore less significant, so we should expect the E-value toincrease. Quantitatively, if the sequence query length were halved (S →S /2m →m /2), the score S would decrease by a factor of 2 ( ) since there are half as many positions at which to accumulate positive match scores. Plugging these into equations for the original query sequence (with score E !) and the half-length query sequence (with score E E !!=!) yields:mn 2!! and E !=mn 2!!/!⇒E !2!=2E !2!/!⇒E !=Thus, the E-value increases essentially exponentially, with an additional decreasing linear factor E !of 2 due to halving m . But this latter effect is much smaller than the exponential 2(!/! ! !).increase resulting from the decreased score.(D)(1 pt.) Returning to the BLAST results from part (A), to what genes and RNA classes do the transcript hits with E-values below 0.05 belong? Does your RNA match the sense or antisense direction of these hits? (Click on the hyperlink of the hit and look at the “Strand” section, which tells you the DNA strand of the Hit/Query.)Of the 2 statistically transcript significant hits at an E-value of 0.05, one matches nucleotides 14-38 of your RNA complementary to (matching the antisense direction of) an mRNA that encodes the phenylalanine hydroxylase (PAH) enzyme. Nucleotides 38-68 of your RNA match the sense direction of Snord100, a C/D Box snoRNA (a type of noncoding RNA that directs posttranscriptional modifications of other RNAs).(E)(2 pts.) After performing an RNA-protein affinity purification (pull-down) from mouse cell lysates followed by mass spectrometry, you determine that your RNA interacts with the product of the ADAR1 gene. What does this enzyme do, and what type of RNA does this enzyme act on? Looking back at the function and strand of the gene hit to the second part of your RNA, state a hypothesis as to how your RNA might function to cause your mouse’s metabolic disorder. (Hint: on the BLAST hit entry corresponding to the mRNA, click on the “Graphics” link to see the hit in red and how your query at the bottom overlaps with it. If ADAR1 acts at the UAU codon, what is the resulting change during translation?)The ADAR1 enzyme catalyzes A-to-I editing, post-transcriptionally deaminating adenosine in double-stranded RNA duplexes, yielding inosine. Since I is interpreted as G during translation, A-to-I changes in protein-coding sequences may lead to codon changes and altered functional properties of the proteins. In addition, A-to-I editing can play important roles in regulating gene expression, such as by altering alternative splicing, miRNA sequences, or miRNA target sites in the mRNA.The PAH gene product is a critical enzyme in phenylalanine metabolism and catalyzes the rate-limiting step in its complete catabolism. Nucleotides 14-38 of your RNA overlap a region of thePAH ORF antisense to the mRNA, including Tyrosine 414 encoded by the codon UAU. Deamination of this adenosine by ADAR would result in the ribosome interpreting a UGU codon, which encodes for the much smaller Cysteine. Thus, your mutant snoRNA provides an RNA duplex for ADAR1 to cause a missense mutation, which could resulting in reduced activity of the PAH enzyme and contribute to your mouse’s metabolic disorder. Indeed, genetic Y414C mutations have been observed in human Phenylketonuria patients, and the mutation has been shown to induce global PAH conformational changes (Gersting et al. Am. Journ. Human Genetics 83 2008 /pmc/articles/PMC2443833/pdf/main.pdf). Note that the RNA found in the wildtype mouse is very similar to the normal Snord100 snoRNA, which directs 2'O-ribose methylation of rRNA and does not affect PAH.Problem 2. Gapped sequence alignment (6 points)In this problem, you will use the algorithms discussed in class to find the optimal alignment for a pair of short peptides.(A) (1 pt.) In order to perform this alignment, you must first choose a scoring matrix. For example, you could use a constant match and mismatch penalty of 1 and -1, respectively, so that−1 otherwise. Is this a good idea? Why or why not? In one sentence,S!"=1 if i = j and S!"=briefly describe how you might obtain a better scoring matrix for protein comparison.No - not all amino acid substitutions are equally (dis)favored. Some changes will more heavily impact protein structure and function than others, and will therefore evolve less frequently, and so they should be scored differently. For example, changing from one medium-sized hydrophobic residue to another (e.g., Val to Ile or Leu) within a signal peptide or transmembrane helix is often tolerated, but changing a hydrophobic to a charged residue could disrupt function in these contexts, and changing a buried medium-sized hydrophobic residue like Val to a much larger residue (e.g., Trp) could disrupt packing. Instead, commonly used scoring matrices are created by comparing related protein sequences and seeing how often evolution has allowed particular substitutions occur - these matrices better capture proteins’ functional constraints than this simple +1/-1 scoring scheme.(B)(1 pt.) You decide to explore more commonly used protein alignment scoring matrices instead. Compare the score for aligning two tryptophans (W) to the score for aligning two alanines (A) in the PAM250 scoring matrix. Both of these alignments are “matches”, so why are these scores so different?W-W pairings have a large positive score, while A-A pairings have a small positive score. This means that tryptophan residues are generally highly conserved, and changes from tryptophan to another amino acid are rare (and therefore generally evolutionarily unfavorable). Conversely, alanine is not as strongly conserved and changes relatively frequently. From a biochemical perspective, this makes sense since alanine is very small and won’t generally have a big impact on protein structure (and is similar to many other nonpolar amino acids), while tryptophan is very big and changing it to almost anything else could dramatically alter protein structure.(C)(2 pts.) Perform a global alignment of the two peptides ATWES and TCAET, using the Needleman-Wunsch algorithm to fill out the alignment matrix below. Use the BLOSUM62 scoring matrix and a linear gap penalty of 2.After filling out the matrix, circle the traceback path and write the final alignment. If there are multiple traceback paths, write out all top-scoring alignments.Using the BLOSUM62 matrix in the textbook or commonly found online:Gap A T W E SGap 0 -2-4-6-8-10T -2031-1-3C -4-211-1-2A -60-1-100E -8-2-1-342T -10-4 3 1 2 5The traceback is highlighted in gray above. The final alignment is:A T W - E S- T C A E TNote: There was a slightly different version of the BLOSUM62 matrix on the lecture slides (the scoring matrix was created from a different set of aligned sequences). This does not change the traceback or final alignment, only a few scores as shown below. Full credit was given for either answer. Using the BLOSUM62 matrix in the lecture slides:Gap A T W E SGap 0 -2-4-6-8-10T -2031-1-3C -4-21310A -61-1131E -8-11-164T -10-3 4 2 4 8(D) (2 pts.) Different scoring matrices and gap penalties can give very different alignment results. Below is the alignment of the peptides from part (C) using the PAM250 scoring matrix (same gap penalty). The traceback path is shaded.Gap A T W E SGap 0 -2 -4 -6 -8 -10T -2 1 1 -1 -3 -5C -4 -1-1 -3 -5 -3A -6 -2 0-2-3 -4E -8 -4 -2 -4 20T -10 -6 -1 -3 0 3What is the resulting alignment?A - T W E ST C A - E TCompare the optimal alignments obtained using the BLOSUM62 and PAM250 scoring matrices. Why are they different?The main reason the alignments are different is because of how strongly the C-W mismatch is penalized under the PAM250 matrix (score = -8), compared to in the BLOSUM62 matrix (score = -2). This means that under BLOSUM62 the C-W mismatch is tolerated without producing a gap, whereas under PAM250 a gap is preferred over the strong -8 penalty. Additionally, under PAM250, A-T pairings are more favorable (score = +1 vs. 0 for BLOSUM62).Problem 3. Sequence similarity search statistics (7 points)You are conducting local nucleotide sequence alignments with your favorite local alignment tool (e.g. BLAST) with match and mismatch scores of +1 and -1 respectively. You align a 100bp query sequence to a 1Mbp genome and find that a 20-nt subsequence from your query is a perfect match.For each of the following cases, calculate the significance of a 20-nt perfect match (assume K = 1 in each case):Note: The Gumbel distribution is continuous, so the P-value for a score x, P(S ≥ x), is equal to the formula P(S > x) on the lecture slides for continuous x since a single point P(S = x) has no probability mass. However, we are applying this continuous distribution to a scoring system that only takes on discrete values, so the P(S = x) values in our scoring system have nonzero mass (a reasonable value for P(S = x) would be CDF(x+1) – CDF(x), where CDF is the cumulative distribution function given on the lecture slides). Thus, our intention was that the P-value is P(S ≥ 20) = P(S > 19), so 19 would be plugged into the Gumbel CDF formula; however, since the lecture slides and the textbook have different wording regarding P(S ≥ x) vs. P(S > x), we will accept P-values with either 19 or 20 used in the Gumbel formula.(A) (2 pts.) Query sequence and genome both have approximately balanced base compositionA=C=G=T=25%).Since every pair of nucleotides occurs with equal probability, the probability of a match (A/A,T/T C/C or G/G) is ¼, and the probability of a mismatch is therefore ¾. So to find λ, we need tosolve !! e!+!!e!!=1, which has solutions λ=0 or ln(3) (by substituting in y=e!). Since λmust be positive, we use λ=ln(3). The score for the perfect 20nt match is x=20, so using the distribution of the scores P(S > x) = 1 - exp[-KMN e!!!], we obtain the P-value:P(S ≥ 20) = P(S > 19) = 1 - exp[-(100)(1000000)e!!"ln(!(0.0283 for )] = 0.0824.x = 20)(B) (1 pt.) Query sequence and genome are both highly A-T rich (A=T=40%, C=G=10%).A/A and T/T matches occur with probability 16/100 while C/C and G/G matches occur with probability 1/100. There are also two mismatches each with probability 16/100 (A/T and T/A) and two with probability 1/100 (C/G and G/C). The remaining 8 pairs are all mismatches with probability 4/100. Overall, the total probability of a match is 34/100 and probability of a mismatch is 66/100. We need to solve (0.34) e!+(0.66)e!!=1, which has nonzero solution λ = 0.6633. The corresponding P-value is:P(S ≥ 20) = P(S > 19) = 1 - exp[-(100)(1000000)e!!"(!.!!"")] ≈1.(also ≈1 for x = 20)(C) (1 pt.) Query is moderately A+T-rich (A = T = 30%, C = G = 20%) but genome is moderately C+G-rich (A = T = 20%, C = G = 30%).In this case, all matches are equiprobable with probability (0.3)(0.2) = 0.06. Therefore the probability of a match is 4(0.06) = 0.24, and the probability of a mismatch is 1-0.24 = 0.76. Solving (0.24)e!+(0.76)e!!=1, we obtain nonzero solution λ = 1.153, and the P-value is:P(S ≥ 20) = P(S > 19) = 1 - exp[-(100)(1000000)e!!"(!.!"#(0.0096 for )] = 0.0301.x = 20)(D)(1 pt.) Briefly explain why the ordering of the P-values from (A) - (C) makes sense.Since in (B) we are searching a highly A-T rich query against a highly A-T rich genome, we expect to see more similarity between the query and the genome by chance than in (A). Therefore, the match becomes much less significant than in (A). When the query is A-T rich and the genome is G-C rich as in (C), however, a match becomes less likely than if both query and genome had equiprobable base compositions as in (A), and so the P-value in (C) is smaller than in (A).(E)(2 pts.) Design a new scoring system for application to searching a 20 nt query of unbiased composition against a highly A+T-rich genome (as in (B) above) that will increase the sensitivity for detection of matches to that genome by drawing lines from each box on the left to its new score in the right box (+1, 0, or -1 for different types of matches/mismatches). What would the P-value of a perfect match to this query (with 5 A’s, 5 C’s, 5 G’s, 5 T’s) be using your new scoring system?Since C/C and G/G matches are unlikely by chance due to their low genome content, observing these matches provides the most evidence of a true alignment; they should therefore be given a score of +1. In contrast, because A/A and T/T matches will occur fairly often simply by chance due to their high genome content, these matches provide less evidence of a true alignment and should be given a score of 0. Mismatches generally provide evidence against a true alignment, so they should be given a score of -1.With a query of unbiased content (A=C=G=T=25%) against the biased genome (A=T=40%,C=G=10%), there is 0.05 total probability of C/C or G/G match (score = +1), 0.2 probability of A/A or T/T match (score = 0), and 0.75 probability of a mismatch (score = -1). The equation! !"" e!+!"!""+!"!""e!!=1 leads to λ =2.7081.For a perfectly matched 20 nt query of unbiased content, there will be 10 matches of score +1 (C/C and G/G) and 10 matches of score 0 (A/A and T/T), for an overall score of +10. The P-value is therefore:P(S ≥ 10) = P(S > 9) = 1 - exp[-(20)(1000000)e!!(!.!"#$(3.4665∗10)] = 5.1988∗10!!!! for x = 10)MIT OpenCourseWare7.91J / 20.490J / 20.390J / 7.36J / 6.802J / 6.874J / HST.506J Foundations of Computational and Systems BiologySpring 2014For information about citing these materials or our Terms of Use, visit: /terms.。
python解多元一次方程组
Python是一种流行的高级编程语言，它可以用于解决各种数学问题，包括解多元一次方程组。
在本文中，我们将探讨如何使用Python来解决多元一次方程组的问题。
让我们来了解一下什么是多元一次方程组。
多元一次方程组是由多个未知数组成的一组方程，每个方程中的未知数都只有一次幂的方程。
解多元一次方程组的过程可以通过消元法、代入法、加减消法等方法来进行，而在Python中，我们可以使用SymPy库来解决这类问题。
接下来，我们将介绍如何使用Python的SymPy库来解决多元一次方程组的问题。
我们需要导入SymPy库，然后定义方程组中的未知数和方程。
1. 导入SymPy库要使用SymPy库，我们首先需要导入它。
在Python中，我们可以使用以下代码来导入SymPy库：```pythonimport sympy as sp```2. 定义未知数和方程接下来，我们需要定义方程组中的未知数和方程。
假设我们有一个包含两个未知数x和y的方程组，我们可以使用以下代码来定义这个方程组：```pythonx, y = sp.symbols('x y')eq1 = sp.Eq(2*x + 3*y, 6)eq2 = sp.Eq(3*x - 2*y, 2)```在这里，我们使用sp.symbols()函数来定义未知数x和y，然后使用sp.Eq()函数来定义方程组中的每个方程。
在这个例子中，我们定义了两个方程eq1和eq2，分别表示2x + 3y = 6和3x - 2y = 2。
3. 解方程组一旦我们定义了未知数和方程，我们就可以使用solve()函数来解决这个方程组。
在Python中，我们可以使用以下代码来解决这个方程组：```pythonsolution = sp.solve((eq1, eq2), (x, y))print(solution)```在这里，我们使用sp.solve()函数来解决方程组(eq1, eq2)，并将未知数(x, y)作为参数传递给solve()函数。
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Harvard University Economics1123
Department of Economics Spring2006
Problem Set1
Smoking and Lung Cancer
Due:Thursday16February.
All of the calculations are automatically done in programs like Stata without you seeing the steps.However it is useful to see them once,so as to appreciate what the computations are as an aid to understanding the output you obtain from the computer.In this problem set you will do the calculations with and without Stata.Thefinal question reminds you to think a bit about the numbers the machine spits out.
You may recall the lines of’Big’Tobacco executives stating that they were not aware of any evidence that definitively linked smoking with lung cancer.The following Table relates per capita cigarette consumption in1930(this is our X variable or regressor)with the subsequent death rates from lung cancer in1950(the Y variable or dependent variable)forfive countries.
ing either a calculator or spreadsheet,compute the following.(if using a spreadsheet attach a printout).
a)¯X,¯Y,i.e.the sample means of both X and Y.
b)s X,s Y,i.e.the sample standard deviations of X and Y.
c)the correlation coefficient between X and Y.
d)ˆβ1,i.e.the slope of the regression line(hint:you can compute this from your previous answers).
e)ˆβ0,the regression intercept term.
f)ˆy i,the predicted values for each country.
g)ˆu i,the estimated residuals for each country.
2.Draw a scatterplot of the data.Put the estimated regression line through the scatterplot. Denote the residuals for each country.
3.Now compute the same statistics using Stata.On the Stata outputfind and label the items in Question1.
STATA:To do this,you need to read the data into Stata.See the Stata hints on the course mands that will be useful are:
list lists the data
summarize reports summary statistics
correlate produces correlation coefficients
regress estimates the regression by OLS
predict computes OLS predicted values and residuals
Note that the Stata output will include numbers we have not yet seen,you can ignore these.
4.Recall that a critical assumption for interpreting the regression was that the residuals are uncorrelated with the X variable.
a)What factors do you think might explain the variation in the residuals(beyond mere ran-domness)?
b)Choose one of your factors in a).Might this factor affect the regression line?How?。