#letters中的元素数量不一定，但肯定是3的倍数

newletter = ["a", "b", "c", "d", "e", "f", "g", "h", "i",]

#想要的结果是每三个元素为一行

print(newletter[0],newletter[1],newletter[2])
print(newletter[3],newletter[4],newletter[5])
print(newletter[6],newletter[7],newletter[8])

a b c
d e f
g h i
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FROM 221.1.219.190

感谢！
for i in range(int(len(newletter) / 3)):
    print(newletter[3*i], newletter[3*i+1], newletter[3*i+2])


我自己的法子太笨了：

col1=[]
col2=[]
col3=[]
for x,y in enumerate(newletter):
    if x%3==0:
        col1.append(y)
        # print(y)
    if x%3==1:
        col2.append(y)
        # print(y)
    if x%3==2:
        col3.append(y)
        # print(y)
    # print()
for i in range(3):
    print(col1[i],col2[i],col3[i])


【 在 Jangbi 的大作中提到: 】
: for i in range(len(newletter) / 3):
:     print(newletter[3*i], newletter[3*i+1], newletter[3*i+2])
--
FROM 111.17.148.156

Got it

for i in range(int(len(newletter) / 3)):
    print("\t".join(newletter[i::int(len(newletter)/3)]))


【 在 annals 的大作中提到: 】
: 还可以试试[::3]这个slice
--
FROM 111.17.148.156