如何在列中打印Python列表？
我想在指定数量的对齐列中显示一个简单列表。例如，以下列表显示在14列中，列之间的空格数在3到1之间变化：如何在列中打印Python列表？,python,list,multiple-columns,Python,List,Multiple Columns,我想在指定数量的对齐列中显示一个简单列表。例如，以下列表显示在14列中，列之间的空格数在3到1之间变化： a = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 3
a = [
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,
15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28,
29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42,
43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56,
57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70,
71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84,
85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98,
99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112,
113, 114, 115, 116, 117, 118, 119, 120
]
for a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, a11, a12, a13, a14 in zip(*[iter(a)] * 14):
print(a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, a11, a12, a13, a14)
1 2 3 4 5 6 7 8 9 10 11 12 13 14
15 16 17 18 19 20 21 22 23 24 25 26 27 28
29 30 31 32 33 34 35 36 37 38 39 40 41 42
43 44 45 46 47 48 49 50 51 52 53 54 55 56
57 58 59 60 61 62 63 64 65 66 67 68 69 70
71 72 73 74 75 76 77 78 79 80 81 82 83 84
85 86 87 88 89 90 91 92 93 94 95 96 97 98
99 100 101 102 103 104 105 106 107 108 109 110 111 112
如何做到这一点？在元素之间添加一个选项卡，然后进行打印似乎对我有用：
for i in range(0, len(a)-1, 14):
print('\t'.join(map(str, a[i:i+14])))
1 2 3 4 5 6 7 8 9 10 11 12 13 14
15 16 17 18 19 20 21 22 23 24 25 26 27 28
29 30 31 32 33 34 35 36 37 38 39 40 41 42
43 44 45 46 47 48 49 50 51 52 53 54 55 56
57 58 59 60 61 62 63 64 65 66 67 68 69 70
71 72 73 74 75 76 77 78 79 80 81 82 83 84
85 86 87 88 89 90 91 92 93 94 95 96 97 98
99 100 101 102 103 104 105 106 107 108 109 110 111 112
113 114 115 116 117 118 119 120
在元素之间添加一个选项卡，然后进行打印似乎对我很有用：
for i in range(0, len(a)-1, 14):
print('\t'.join(map(str, a[i:i+14])))
1 2 3 4 5 6 7 8 9 10 11 12 13 14
15 16 17 18 19 20 21 22 23 24 25 26 27 28
29 30 31 32 33 34 35 36 37 38 39 40 41 42
43 44 45 46 47 48 49 50 51 52 53 54 55 56
57 58 59 60 61 62 63 64 65 66 67 68 69 70
71 72 73 74 75 76 77 78 79 80 81 82 83 84
85 86 87 88 89 90 91 92 93 94 95 96 97 98
99 100 101 102 103 104 105 106 107 108 109 110 111 112
113 114 115 116 117 118 119 120
您可以使用一些字符串格式：
from itertools import izip_longest
width = 14
for lst in izip_longest(*[iter(a)]*width, fillvalue=''):
print(('{:3} '*width).format(*lst))
zip您可以使用一些字符串格式：
from itertools import izip_longest
width = 14
for lst in izip_longest(*[iter(a)]*width, fillvalue=''):
print(('{:3} '*width).format(*lst))
zip这不是一个漂亮的解决方案，但您可以根据整数的字符串长度添加空格：
a = [
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,
15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28,
29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42,
43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56,
57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70,
71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84,
85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98,
99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112,
113, 114, 115, 116, 117, 118, 119, 120
]
maxLen = len(str(max(a)))
a = [str(A)+" "*(maxLen-len(str(A))) for A in a]
for a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, a11, a12, a13, a14 in zip(*[iter(a)] * 14):
print(a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, a11, a12, a13, a14)
1 2 3 4 5 6 7 8 9 10 11 12 13 14
15 16 17 18 19 20 21 22 23 24 25 26 27 28
29 30 31 32 33 34 35 36 37 38 39 40 41 42
43 44 45 46 47 48 49 50 51 52 53 54 55 56
57 58 59 60 61 62 63 64 65 66 67 68 69 70
71 72 73 74 75 76 77 78 79 80 81 82 83 84
85 86 87 88 89 90 91 92 93 94 95 96 97 98
99 100 101 102 103 104 105 106 107 108 109 110 111 112
这不是一个漂亮的解决方案，但您可以根据整数的字符串长度添加空格：
a = [
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,
15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28,
29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42,
43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56,
57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70,
71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84,
85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98,
99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112,
113, 114, 115, 116, 117, 118, 119, 120
]
maxLen = len(str(max(a)))
a = [str(A)+" "*(maxLen-len(str(A))) for A in a]
for a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, a11, a12, a13, a14 in zip(*[iter(a)] * 14):
print(a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, a11, a12, a13, a14)
1 2 3 4 5 6 7 8 9 10 11 12 13 14
15 16 17 18 19 20 21 22 23 24 25 26 27 28
29 30 31 32 33 34 35 36 37 38 39 40 41 42
43 44 45 46 47 48 49 50 51 52 53 54 55 56
57 58 59 60 61 62 63 64 65 66 67 68 69 70
71 72 73 74 75 76 77 78 79 80 81 82 83 84
85 86 87 88 89 90 91 92 93 94 95 96 97 98
99 100 101 102 103 104 105 106 107 108 109 110 111 112
另一个没有itertools的简单解决方案：
i=0
while i<len(a):
for j in range(i,min(i+14,len(a))):
print '%4d' % (a[j]),
print
i+=14
i=0
而我另一个不含itertools的简单解决方案：
i=0
while i<len(a):
for j in range(i,min(i+14,len(a))):
print '%4d' % (a[j]),
print
i+=14
i=0
当i时，下面计算出填充每个条目所需的最大位数。因此，在这种情况下需要3
。如果将1000
添加到列表中，则该列表将变为4
from itertools import izip_longest
import math
a = [
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,
15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28,
29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42,
43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56,
57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70,
71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84,
85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98,
99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112,
113, 114, 115, 116, 117, 118, 119, 120
]
pad = '{{:{}}}'.format(int(math.ceil(math.log(max(a)+1) / math.log(10))))
for row in izip_longest(*[iter(a)] * 14, fillvalue=''):
print ' '.join(pad.format(v) for v in row)
izip_longest
与填充值一起用于填充任何剩余条目。这将为您提供以下输出：
1234567891011214
15 16 17 18 19 20 21 22 23 24 25 26 27 28
29 30 31 32 33 34 35 36 37 38 39 40 41 42
43 44 45 46 47 48 49 50 51 52 53 54 55 56
57 58 59 60 61 62 63 64 65 66 67 68 69 70
71 72 73 74 75 76 77 78 79 80 81 82 83 84
85 86 87 88 89 90 91 92 93 94 95 96 97 98
99 100 101 102 103 104 105 106 107 108 109 110 111 112
113 114 115 116 117 118 119 120
例如，如果您的数字列表只上升到50
，它将自动调整如下：
for a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, a11, a12, a13, a14 in zip(*[iter(a)] * 14):
print(a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, a11, a12, a13, a14)
1234567891011214
15 16 17 18 19 20 21 22 23 24 25 26 27 28
29 30 31 32 33 34 35 36 37 38 39 40 41 42
43 44 45 46 47 48 49 50
下面计算出填充每个条目所需的最大位数。因此，在这种情况下需要3
。如果将1000
添加到列表中，则该列表将变为4
from itertools import izip_longest
import math
a = [
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,
15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28,
29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42,
43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56,
57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70,
71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84,
85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98,
99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112,
113, 114, 115, 116, 117, 118, 119, 120
]
pad = '{{:{}}}'.format(int(math.ceil(math.log(max(a)+1) / math.log(10))))
for row in izip_longest(*[iter(a)] * 14, fillvalue=''):
print ' '.join(pad.format(v) for v in row)
izip_longest
与填充值一起用于填充任何剩余条目。这将为您提供以下输出：
1234567891011214
15 16 17 18 19 20 21 22 23 24 25 26 27 28
29 30 31 32 33 34 35 36 37 38 39 40 41 42
43 44 45 46 47 48 49 50 51 52 53 54 55 56
57 58 59 60 61 62 63 64 65 66 67 68 69 70
71 72 73 74 75 76 77 78 79 80 81 82 83 84
85 86 87 88 89 90 91 92 93 94 95 96 97 98
99 100 101 102 103 104 105 106 107 108 109 110 111 112
113 114 115 116 117 118 119 120
例如，如果您的数字列表只上升到50
，它将自动调整如下：
for a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, a11, a12, a13, a14 in zip(*[iter(a)] * 14):
print(a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, a11, a12, a13, a14)
1234567891011214
15 16 17 18 19 20 21 22 23 24 25 26 27 28
29 30 31 32 33 34 35 36 37 38 39 40 41 42
43 44 45 46 47 48 49 50
这称为漂亮的打印。如果python没有一个函数可以为您完成这项工作，那么您需要在较小的数字上填充空格，以便所有内容都对齐。这称为“漂亮打印”。如果python没有一个函数可以为您实现这一点，那么您需要用空格填充较小的数字，以便所有内容都对齐。打印后列表将变为空。打印后列表将变为空。