我有这个列表:
a = [(741.0, 0), (743.0, 0), (3386.0, 0), (284577.0, 0), (290611.0, 0), (300889.0, 3), (305256.0, 0), (917458.0, 0), (917905.0, 0), (917906.0, 0), (922187.0, 0), (925852.0, 0), (1260021.0, 0), (1377096.0, 0), (1524210.0, 0), (1680657.0, 0), (1692571.0, 0), (1692645.0, 0), (1692647.0, 0), (1713958.0, 0), (1801008.0, 0), (1818975.0, 0), (1858888.0, 0), (1880544.0, 0), (1880898.0, 0), (1880899.0, 0), (1880900.0, 0), (1881062.0, 0), (1881073.0, 0), (1881240.0, 0), (1881433.0, 0), (1881434.0, 0), (1881435.0, 0), (1881436.0, 0), (1881438.0, 0), (1958358.0, 0), (1958478.0, 0), (1958479.0, 0), (1958481.0, 0), (1967310.0, 0)]
我想创建 2 个列表...第一个列表将包含上面每个列表中的第一个值，b[0] = [741.0,743.0,3386.0,....] 并且第二个将具有第二个值，a = [(j,k,l),(j1,k1,l1),...]。我想用 for 来做这件事，这样如果 a = [(j,k,l),(j1,k1,l1),...] 我会创建 3 个列表而不是 2.. .我有这段代码，但它不能按我想要的方式工作:
代码示例:
b = []
for i in range(N):
b.append([])
for j in range(R):
b[i].append([])
for k in range(0,M,2):
b[i][j].append(j - k)
print b
N = 2，R = 40，M = 80 (2x40)
最佳答案
更Pythonic的方法在于使用列表理解:
for i in range(len(a[0])):
b.append([elt[i] for elt in a])
关于python - 如何将二维列表中的值 append 到另一个列表并打印它，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/37298085/