處理十六進制數時遇到問題。str (hex ())從檔案運行時，后面的零0x...消失。
在入口：
0x0000000000000000000000000000000000000000000000000000000000000001f01f80f12f7cf16638f7a8074d46fe2f421a73432b1441a01ed3dd883c68acad
0x00000000000000000000000000000000000000000000000000000000000000029a799033fc54073346f870c15c9836f6b2e9eccdb85f09d29a8ddc90dc3a8ef1
0x00000000000000000000000000000000000000000000000000000000000000033e561483073e429ec25c09c99de2a81d5a34a539ad2dbb688af6b6f5f24936a4
退出時：
0x1f01f80f12f7cf16638f7a8074d46fe2f421a73432b1441a01ed3dd883c68acad
0x29a799033fc54073346f870c15c9836f6b2e9eccdb85f09d29a8ddc90dc3a8ef1
0x33e561483073e429ec25c09c99de2a81d5a34a539ad2dbb688af6b6f5f24936a4
代碼：
with open("data.txt", "r") as file:
for line in file:
L = int(line, 0)
R = str(hex(L))
print(R)
代碼中需要修復什么？我需要一種大小的數字并且不會丟失零。
uj5u.com熱心網友回復：
使用字串格式：
#意味著放在0x前面的十六進制數字。0130表示欄位長度為 130 個字符，前導零表示用零填充。x表示十六進制（小寫 af）。
line = '0x0000000000000000000000000000000000000000000000000000000000000001f01f80f12f7cf16638f7a8074d46fe2f421a73432b1441a01ed3dd883c68acad'
print(line) # as read from file
integer = int(line, 0)
formatted = f'{integer:#0130x}'
print(formatted)
print(formatted == line) # check that original and re-formatted are the same
輸出：
0x0000000000000000000000000000000000000000000000000000000000000001f01f80f12f7cf16638f7a8074d46fe2f421a73432b1441a01ed3dd883c68acad
0x0000000000000000000000000000000000000000000000000000000000000001f01f80f12f7cf16638f7a8074d46fe2f421a73432b1441a01ed3dd883c68acad
True
轉載請註明出處，本文鏈接：https://www.uj5u.com/qita/314370.html
上一篇：根據規則對字串串列求和
下一篇：Python中的串列切片與索引？