python 1,2,3...
不用科学计数法打印:
print("{:.7f}".format(6.87e-5))
插入一段list
a= [ 1, 2, 3 ]
a[1:1] = [ 4,5,6 ]
【0】out: [1, 4, 5, 6, 2, 3]
找出可能原始数据的无效值
np.any(np.isnan(val))
np.any(np.isnan(val))
print("{:.7f}".format(6.87e-5))
a= [ 1, 2, 3 ]
a[1:1] = [ 4,5,6 ]
【0】out: [1, 4, 5, 6, 2, 3]