lstA 看起來像這樣：
["y1","y2","y3"]
lstB 看起來像這樣：
["xx5, folder, 20-1-1", "xx6, appPath, 20-1-1", "xx7, Resolve, 23-1-4"]
所有專案都是字串。串列的長度相同。
我想遍歷 lstB 的每一項，并分別用lstA中的元素替換之前的第一項,。
最后，lstB應該如下所示：
["y1, folder, 20-1-1", "y2, appPath, 20-1-1", "y3, Resolve, 23-1-4"]
首先，我可以遍歷 lstB，將字串拆分為 ，以獲得第一項的值（例如“xx5”）。然后，我可以用 lstA 中的相應元素替換它。但是，我不確定如何替換，因為 lstA 是獨立的。
for i in lstB:
toBeReplaced = i.split(',')[0]
totalName = i.replace(toBeReplaced, lstA[i])
uj5u.com熱心網友回復：
如果您不熟悉編程，則方法非常簡單。
- 代碼：
a = ["y1","y2","y3"]
b = ["xx5, folder, 20-1-1", "xx6, appPath, 20-1-1", "xx7, Resolve, 23-1-4"]
List = []
for i in range(len(a)):
temp = b[i].split(',')
temp[0] = a[i]
List.append((',').join(temp))
- 輸出：
['y1, folder, 20-1-1', 'y2, appPath, 20-1-1', 'y3, Resolve, 23-1-4']
uj5u.com熱心網友回復：
拆分和連接兩個串列的專案，可以在一個簡短的串列理解中完成：
lstA = ["y1","y2","y3"]
lstB = ["xx5, folder, 20-1-1", "xx6, appPath, 20-1-1", "xx7, Resolve, 23-1-4"]
lstB = [','.join((a, *b.split(',')[1:])) for a, b in zip(lstA, lstB)]
uj5u.com熱心網友回復：
您可以使用索引變數遍歷第二個串列，并使用字串切片運算子i將前 3 個字符替換為第一個串列中的相應值：[:]
l = ["y1","y2","y3"]
l1 = ["xx5, folder, 20-1-1", "xx6, appPath, 20-1-1", "xx7, Resolve, 23-1-4"]
for i in range(len(l1)):
l1[i] = (l[i] l1[i][3:])
print(l1)
輸出：
['y1, folder, 20-1-1', 'y2, appPath, 20-1-1', 'y3, Resolve, 23-1-4']
uj5u.com熱心網友回復：
可以很容易地使用zip
a=["y1","y2","y3"]
b=["xx5, folder, 20-1-1", "xx6, appPath, 20-1-1", "xx7, Resolve, 23-1-4"]
print(b)
for i,(a_,b_) in enumerate(zip(a,b)):
b[i]=b_.replace(b_.split(",")[0], a_)
print(b)
輸出：
before=['xx5, folder, 20-1-1', 'xx6, appPath, 20-1-1', 'xx7, Resolve, 23-1-4']
after=['y1, folder, 20-1-1', 'y2, appPath, 20-1-1', 'y3, Resolve, 23-1-4']
uj5u.com熱心網友回復：
使用串列推導構造新串列
- 壓縮 2 個串列并遍歷它們。
- 使用
split()at,并限制 maxsplit=1 - 使用 f-string 創建新元素
代碼：
spam = ["y1","y2","y3"]
eggs = ["xx5, folder, 20-1-1", "xx6, appPath, 20-1-1", "xx7, Resolve, 23-1-4"]
print([f'{item1}, {item2.split(",", maxsplit=1)[-1]}' for item1, item2 in zip(spam, eggs)])
輸出
['y1, folder, 20-1-1', 'y2, appPath, 20-1-1', 'y3, Resolve, 23-1-4']
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標籤：Python python-3.x 列表 麻木的 字符串替换