非常简单的代码,我有一个带有两个值的字符extraAccessions,一个字符串和一个列表.我想循环遍历extraAccessions(在这个例子中只有一个)并使用元组中的第一个和第二个值.
extraAccessions=('MS:1000505',['value','unitName'])
for accession, fieldIdentifiers in extraAccessions:
[do something]
但是,这给了
ValueError: too many values to unpack
当我做
print (extraAccessions)
我明白了
('MS:1000505', ['value', 'unitName'])
这对我来说似乎有两个价值,正是所要求的
for accession, fieldIdentifiers in extraAccessions:
所以我不明白为什么我会收到这个错误.
编辑:
而当我这样做
for accession in extraAccessions:
print accession
我得到第一个元素MS:1000505
我有一个数据框,其中的列包含可变数量的数字和可变数量的NA.数据框如下所示:
V1 V2 V3 V4 V5 V6
1 0 11 4 0 0 10
2 0 17 3 0 2 2
3 NA 0 4 0 1 9
4 NA 12 NA 1 1 0
<snip>
743 NA NA NA NA 8 NA
744 NA NA NA NA 0 NA
我想制作一个箱形图,但是当我这样做的时候
boxplot(dataframe)
我收到了错误
adding class "factor" to an invalid object
当我做
lapply(dataframe,class)
我得到以下输出:
$V1
[1] "factor"
$V2
[1] "factor"
<snip>
$V6
[1] "factor"
那么如何更改我的数据框以便将列视为数字？
我有一个SQL查询,我检查一个值是否在表的最大值和最小值之间.我现在实现如下:
SELECT spectrum_id, feature_table_id
FROM 'spectrum', 'feature'
WHERE `spectrum`.msrun_msrun_id = 1
AND `feature`.msrun_msrun_id = 1
AND (SELECT min(rt) FROM `convexhull` WHERE `convexhull`.feature_feature_table_id = `feature`.feature_table_id) <= scan_start_time
AND scan_start_time <= (SELECT max(rt) FROM `convexhull` WHERE 'convexhull'.feature_feature_table_id = 'feature'.feature_table_id)
AND (SELECT min(mz) FROM `convexhull` WHERE `convexhull`.feature_feature_table_id = `feature`.feature_table_id) <= base_peak_mz
AND base_peak_mz <= (SELECT max(mz) FROM `convexhull` WHERE `convexhull`.feature_feature_table_id = `feature`feature_table_id)
这运行得非常慢,因为每次运行此查询时我都会从convexhull中选择4次,所以我尝试使用内部连接来改进它:
SELECT spectrum_id, feature_table_id
FROM 'spectrum', 'feature'
INNER JOIN `convexhull` ON `convexhull`.feature_feature_table_id = `feature`.feature_table_id
WHERE `spectrum`.msrun_msrun_id = ? "+
AND `feature`.msrun_msrun_id …如何更改此代码,以便每60秒调用一次doWork().目前,它正在循环,并在60秒后停止.想要完全相反.
from twisted.internet import task
from twisted.internet import reactor
import twapi
timeout = 200.0 # Sixty seconds
def doWork():
#do work here
twapi.main('1')
pass
l = task.LoopingCall(doWork)
l.start(timeout) # call every sixty seconds
reactor.run()
我有以下递归函数
def collatz(n,seq=[]):
seq.append(n)
if n == 1:
return seq
if n%2 == 0:
return collatz(n/2, seq)
else:
return collatz((3*n)+1, seq)
当我多次运行此函数时,seq仍然包含先前运行的值:
>>> collatz(1)
[1] # correct
>>> collatz(2)
[1, 2, 1] # should be [2,1]
>>> collatz(3)
[1, 2, 1, 3, 10, 5, 16, 8, 4, 2, 1] # should be [3, 10, 5, 16, 8, 4, 2, 1]
防止这种情况的方法是seq在调用时给出一个空列表collatz:
>>> collatz(3,seq=[])
[3, 10, 5, 16, 8, 4, 2, 1]
有没有办法以collatz不同的方式写,以便我可以collatz(3)代替 …
我有两种类型的数据,我想做成对比较:假(控件)和测量,
sham1 <- c('Sham1.r1', 'Sham1.r2', 'Sham1.r3')
sham2 <- c('Sham2.r1', 'Sham2.r2', 'Sham2.r3')
shams <- list(sham1, sham2)
day14 <- c('T14d.r1', 'T14d.r2', 'T14d.r3')
day90 <- c('T90d.r1', 'T90d.r2', 'T90d.r3')
measurements <- list(hour4, day1, day3, day7, day14, day90)
我想得到以下内容:
(('Sham1.r1', 'Sham1.r2', 'Sham1.r3'), ('T14d.r1', 'T14d.r2', 'T14d.r3')), (('Sham1.r1', 'Sham1.r2', 'Sham1.r3'), ('T90d.r1', 'T90d.r2', 'T90d.r3')), (('Sham2.r1', 'Sham2.r2', 'Sham2.r3'), ('T14d.r1', 'T14d.r2', 'T14d.r3')), (('Sham2.r1', 'Sham2.r2', 'Sham2.r3'), ('T90d.r1', 'T90d.r2', 'T90d.r3'))
我想要的是:
list(list(c("Sham1.r1", "Sham1.r2", "Sham1.r3"), c("T14d.r1",
"T14d.r2", "T14d.r3")), list(c("Sham1.r1", "Sham1.r2", "Sham1.r3"
), c("T90d.r1", "T90d.r2", "T90d.r3")), list(c("Sham2.r1", "Sham2.r2",
"Sham2.r3"), c("T14d.r1", "T14d.r2", "T14d.r3")), list(c("Sham2.r1",
"Sham2.r2", …我有一个命名列表列表
t <- list(list(a=1,b=2,c=3), list(a=3, b=2, c=1), list(a=2, b=3, c=1))
[[1]]
[[1]]$a
[1] 1
[[1]]$b
[1] 2
[[1]]$c
[1] 3
[[2]]
[[2]]$a
[1] 3
[[2]]$b
[1] 2
[[2]]$c
[1] 1
[[3]]
[[3]]$a
[1] 2
[[3]]$b
[1] 3
[[3]]$c
[1] 1
我希望得到一个命名列表或数据框,就像这样
$a
[1] 1 3 2
$b
[1] 2 2 3
$c
[1] 3 1 1
目前,我循环遍历命名列表列表,将每个值添加到新的命名列表.我怎么能在一行中做到这一点？
我有一个看起来像这样的元组列表:
[(334.99972431901307, 0.0), (335.00088248902574, 0.0), (335.0020406650446, 0.0), (335.0031988470696, 66.83868408203125), (335.00435703510072, 252.91905212402344), (335.0055152291381, 341.447509765625), (335.00667342918183, 282.1964111328125), (335.0078316352317, 125.92335510253906), (335.00898996725408, 0.0), (335.01014818531672, 0.0)]
该列表的长度为16665,并按每个元组的第一个元素排序.我想基于第一个元组值从列表中提取特定范围的元组.目前我这样做:
def getSpectra_mzWindow(self, mzStart, mzEnd):
for spectrum in self.mzmlInstance:
# loop through all the peaks
for peak in spectrum.peaks:
# it's ordered, so when peak[0] > mzEnd it can stop
if float(peak[0]) > mzEnd:
break
if mzStart <= float(peak[0]) <= mzEnd:
yield spectrum, peak
然而,这非常缓慢.既然我知道它是按照第一个值排序的,那么有没有比循环遍历整个列表更快的方法呢？我正在考虑实现二进制搜索,但是已经有一个已经对已排序元组列表进行优化的库了吗？
我正在尝试编译其他人的C++代码.我自己有0个C++经验.我在MAC上使用g ++来编译我收到的一个.cpp文件.当我这样做时,g++ main.cpp我得到一个未定义的符号错误.当谷歌搜索答案似乎是文件之间的错误链接,但我不知道如何链接文件.如何让文件编译？我粘贴了下面的完整代码.
Undefined symbols:
"initializeFitness()", referenced from:
runEvolution() in ccZXBTDH.o
"Grid::GetNeighbourhood(int, int, std::vector<Agent**, std::allocator<Agent**> >&)", referenced from:
runEvolution() in ccZXBTDH.o
runEvolution() in ccZXBTDH.o
runEvolution() in ccZXBTDH.o
"Grid::Grid()", referenced from:
runEvolution() in ccZXBTDH.o
"Reaper::GetAgentToKill()", referenced from:
runEvolution() in ccZXBTDH.o
"Cupid::GetRandomBreeder()", referenced from:
runEvolution() in ccZXBTDH.o
"Reaper::Select()", referenced from:
runEvolution() in ccZXBTDH.o
"Grid::~Grid()", referenced from:
runEvolution() in ccZXBTDH.o
runEvolution() in ccZXBTDH.o
"Breeder::Breed(Agent**, Agent**)", referenced from:
runEvolution() in ccZXBTDH.o
"Cupid::Select()", referenced from:
runEvolution() in ccZXBTDH.o
"Cupid::GetEmptyCell()", referenced from:
runEvolution() in …我正在尝试打印以下代码返回的值:
Agent** Grid::GetAgent(int x, int y)
{
return &agents[x][y];
}
它返回一个双指针,并打印
std::cout << *grid.GetAgent(j, k) << endl;
给出了一个内存位置,但是当我尝试时
std::cout << **grid.GetAgent(j, k) << endl;
我收到了错误
main.cpp:53: error: no match for ‘operator<<’ in ‘std::cout << * * grid.Grid::GetAgent(j, k)’
如何从*grid.GetAgent(j,k)打印值？
以下是Agent.h
#ifndef AGENT_H
#define AGENT_H
enum AgentType { candidateSolution, cupid, reaper, breeder};
class Agent
{
public:
Agent(void);
~Agent(void);
double GetFitness();
int GetAge();
void IncreaseAge();
AgentType GetType();
virtual void RandomizeGenome() = 0;
protected:
double m_fitness;
AgentType m_type;
private:
int m_age;
};
#endif …我有一个文件夹列表,通过
import glob
list_of_folders = glob.glob('path_to_folder/*')
for folder in list_of_folders:
现在,我想对list_of_folders排序文件夹中的文件数量.就像是
list_of_folders.sort(key= lambda f: len(x) for x in glob.glob(f))
但无论我尝试什么,我都无法得到一个有效的lambda表达式.如何对文件夹中文件数量的文件夹列表进行排序？