弄了一下午还是没想明白:
eg:
d1 = [{"eslid":"5A-15-D4-99","nw1":"51-02-01-66","nw3":"50",'rfpower': '50', 'netid': '65', 'apid': 1, 'version': '51', 'battery': '30', 'reverse': '0'}] * 10
d2 = [{"eslid":"5A-15-D4-99","nw1":"51-02-01-66","nw3":"50",'rfpower': '50', 'netid': '65', 'apid': 1, 'version': '51', 'battery': '30', 'reverse': '0'}, {"eslid":"5A-15-D4-99","nw1":"51-02-01-66","nw3":"50",'rfpower': '50', 'netid': '65', 'apid': 1, 'version': '51', 'battery': '30', 'reverse': '0'}, {"eslid":"5A-15-D4-99","nw1":"51-02-01-66","nw3":"50",'rfpower': '50', 'netid': '65', 'apid': 1, 'version': '51', 'battery': '30', 'reverse': '0'}, {"eslid":"5A-15-D4-99","nw1":"51-02-01-66","nw3":"50",'rfpower': '50', 'netid': '65', 'apid': 1, 'version': '51', 'battery': '30', 'reverse': '0'}, {"eslid":"5A-15-D4-99","nw1":"51-02-01-66","nw3":"50",'rfpower': '50', 'netid': '65', 'apid': 1, 'version': '51', 'battery': '30', 'reverse': '0'}, {"eslid":"5A-15-D4-99","nw1":"51-02-01-66","nw3":"50",'rfpower': '50', 'netid': '65', 'apid': 1, 'version': '51', 'battery': '30', 'reverse': '0'}, {"eslid":"5A-15-D4-99","nw1":"51-02-01-66","nw3":"50",'rfpower': '50', 'netid': '65', 'apid': 1, 'version': '51', 'battery': '30', 'reverse': '0'}, {"eslid":"5A-15-D4-99","nw1":"51-02-01-66","nw3":"50",'rfpower': '50', 'netid': '65', 'apid': 1, 'version': '51', 'battery': '30', 'reverse': '0'}, {"eslid":"5A-15-D4-99","nw1":"51-02-01-66","nw3":"50",'rfpower': '50', 'netid': '65', 'apid': 1, 'version': '51', 'battery': '30', 'reverse': '0'}, {"eslid":"5A-15-D4-99","nw1":"51-02-01-66","nw3":"50",'rfpower': '50', 'netid': '65', 'apid': 1, 'version': '51', 'battery': '30', 'reverse': '0'}]
#分别对d1和d2进行操作
for i in range(10):
d1[i]['eslid'] = L[i][0]
d1[i]['nw1'] = L[i][1]
d1[i]['nw3'] = L[i][2]
#然后分别打印d1与d2
发现d1是最后更新的值
d2的值是正常更新的
为什么呢?