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Python zip(*list)[num]为什么会把列表的一列提取出来

zip()是Python的一个内建函数,它接受一系列可迭代的对象作为参数,将对象中对应的元素打包成一个个tuple(元组),第0个元组对应于所有参数的第0个元素,第1个元组对应于所有参数的第1个元素,依此类推,然后返回由这些tuples组成的list(列表...

: 把列表转化为字典就可以了,先声明个字典dict={} 然后dict[001]="老鼠",dict[002]="鸡"就可以了

>>> arr[[10, 11, 12, 13, 14, 15, 16, 17, 18, 19], [11, 12, 13, 14, 15, 16, 17, 18, 19, 20], [12, 13, 14, 15, 16, 17, 18, 19, 20, 21], [13, 14, 15, 16, 17, 18, 19, 20, 21, 22], [14, 15, 16, 17, 18, 19, 20, 21, 22, 23], [15, 16, ...

python3.x result=[] with open('a.txt','r') as f: for line in f: result.append(list(map(float,line.split(',')))) print(result) python2.7 result=[] with open('a.txt','r') as f: for line in f: result.append(map(float,line.split(',...

#!/bin/python a = [] for x in [1,2,3,4,5,6,7,8,9,10,11]: a.append(x) print a for代表循环取数。 append()函数代表将值插入到列表最后。

##文件中内容cat a.txt (('aa', 6L), [('n', '3CE'), ('c', '48'), ('c', 'ff')])(('bb', 5L), [('n', '4E5'), ('c', '28'), ('c', '2'), ('c', '8')])(('cc', 2L), [('n', '5DC'), ('c', '108'), ('c', '4'), ('c', '2'), ('c', '4')])#python...

import xlrddata = xlrd.open_workbook('excelFile.xls')table = data.sheet_by_index(0) #通过索引顺序获取,0表示第一张表data = [table.cell(i,ord('G')-ord('A')).value for i in range(1, 90)]

Python 3.3.3 (v3.3.3:c3896275c0f6, Nov 18 2013, 21:18:40) [MSC v.1600 32 bit (Intel)] on win32 Type "copyright", "credits" or "license()" for more information. >>> D= {'a':1,'b':2,'c':3} >>> list(D.keys()) ['b', 'c', 'a'] >>> D...

单纯从数据结构上来讲的话,字典的数据结构是散列,也就是哈希表,它是无序的,而列表的数据结构是链表,是有序的,所以想把字典转换成列表并且保持顺序不变,是不可行的。 题主可以使用OrderedDict(Python 2.7+)有序字典,OrderedDict是dict...

def loadDataSet(filename): dataMat=[] fr=open(filename) for line in fr.readlines(): line = line.replace('"','') curLine=line.strip().split('\t') aa = [float(i) for i in curLine] dataMat.append(aa) return dataMatdataMat=loadData...

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