如何从长十进制值列表中选择最后 3 位数字

Question

下面是十进制数列表；

000000 =  0.0218354767535

000001 =  0.0218265654136

000002 = 0.0218184623573

000003 =  0.021811165579

000004 =  0.0218046731276

000005 =  0.0217989831063

000006 =  0.0217940936718

000007 =  0.0217900030345

000008 =  0.0217867094577

000009 =  0.0217842112574

000010 =  0.021782506802

我想获取给定列表中每个十进制数的最后三位数字。以下是我在这个例子中的需要。

对于可能的解决方案有什么建议吗？

Answer 1

对于每一行，您只能像这样索引最后三个元素：

with open('filename') as infile:
    for line in infile:
        print line.rstrip()[-3:]

Answer 2

假设给定的列表是一个字符串列表。在这种情况下，请使用此代码。

l1 = ["000000 =  0.0218354767535",
"000001 =  0.0218265654136",
"000002 = 0.0218184623573",
"000003 =  0.021811165579",
"000004 =  0.0218046731276",
"000005 =  0.0217989831063",
"000006 =  0.0217940936718",
"000007 =  0.0217900030345",
"000008 =  0.0217867094577",
"000009 =  0.0217842112574",
"000010 =  0.021782506802"]

for each_item in l1:
    print each_item[-3:]

收到澄清后修改答案

result_list = [] # This would store the end result with int
# The below list is the input list with floats
l1 = [0.0218354767535,
0.0218265654136,
0.0218184623573,
0.021811165579,
0.0218046731276,
0.0217989831063,
0.0217940936718,
0.0217900030345,
0.0217867094577,
0.0217842112574,
0.021782506802]

#For each float in the list, find the decimal part, convert to string, extract last 3, convert to int and append to result list
for each_item in l1:
    result_list.append(int(str(each_item-int(each_item))[-3:]))

print result_list

输出

[535, 136, 573, 579, 276, 63, 718, 345, 577, 574, 802]

如何从长十进制值列表中选择最后 3 位数字

how to pick last 3 digits from the list of long decimal values

math

list

sequence

python-2.7