我想尝试另一个示例，以打印2D列表的索引。示例：
a = [["text", "man","chest","funny"],["cruel", "just","for","testing"],["I", "take","this","for"],["learning", "purpose","only","please"] ]
b = [["text", "funny"], ["cruel"],["I", "take"], ["for","learning", "purpose"]]
ba = ["text", "funny", "purpose"]
我的代码是这样的：
store_x = []
for x in b:
for i,xx in enumerate(x):
if xx in ba:
store_x.append(i)
print(store_x)
dic2 = []
for x,y in zip(store_x,a):
result = []
for u in str(x):
dic2.append(y[int(u)])
print(dic2)
当前输出：
[0, 1, 2]
['text', 'just', 'this']
预期输出：
[0][0,1],[3][2] # {b} based on {ba} Not sure whether this is how the index #should be look like for 2D
[["text","man"],["only"]] # store_x based on {a}
我想首先基于{ba}找到{b}的索引值，然后从该索引值中，通过使用存储在{store_x}中的索引，将其用于从{a}打印该值
新功能：
当我们处理{ba}元素时，这是另一个问题。例如
a = [["text", "funny"], ["cruel"],["I", "take"], ["for","learning", "purpose"]]
b = [["text", "funny"], ["cruel"],["I", "take"], ["for","learning", "purpose"]]
ba = ["funny","text", "take","new","I", "purpose", "learning", "cruel"]
然后我当前使用的代码是：
store_x = {}
for ex, x in enumerate(b):
row = []
for i, xx in enumerate(x):
if xx in ba:
row.append(i)
if row:
store_x[ex] = row
print(store_x)
order = {e: ii for ii, e in enumerate(ba)}
dic2 = []
for i, x in store_x.items():
row = []
if x:
for ex in x:
dic2.append(a[i][ex])
if dic2:
dic2.sort(key=lambda e: order.get(e, len(ba)))
print(dic2)
store_x值似乎是：
{0: [0, 1], 1: [0], 2: [0, 1], 3: [1, 2]}
不是我所期望的：
{0: [1, 0], 2: [1,0], 3: [2, 1], 1: [0]}
当前dic2 ：（这是正确的，但store_x值不是我期望的值）
['funny', 'text', 'take', 'I', 'purpose', 'learning', 'cruel']
更新！
我尝试使用我的新代码：
s = []
for ex, x in enumerate(ba):
store_x = {}
for i, xx in enumerate(b):
if x in xx:
store_x[i] = [xx.index(x)]
s.append(store_x)
print(s)
dic2 = []
for x in s:
for i, xx in x.items():
row = []
if xx:
for u in xx:
dic2.append(a[i][u])
print(dic2)
输出为：
[{0: [1]}, {0: [0]}, {2: [1]}, {}, {2: [0]}, {3: [2]}, {3: [1]}, {1: [0]}]
['funny', 'text', 'take', 'I', 'purpose', 'learning', 'cruel']
但这仍然不是我期望的输出：
我的预期输出：
{0: [1, 0], 2: [1,0], 3: [2, 1], 1: [0]}
有人可以帮我这个忙吗？
高于全部！
发现了另一个问题。
代码：
a = [["text", "funny"], ["cruel"],["I", "take"], ["for","learning", "purpose"], ["learning", "to", "learning", "when", "I", "have", "time", "for", "learning"]]
b = [["text", "funny"], ["cruel"],["I", "take"], ["for","learning", "purpose"], ["learning", "to", "learning", "when", "I", "have", "time", "for", "learning"]]
ba = ["funny","text","learning", "take","new","I", "purpose", "cruel"]
从Daniel Mesejo（好家伙）那里获得解决方案时。
输出变为：
{0: [1, 0], 1: [0], 2: [1, 0], 3: [1, 2], 4: [0, 2, 7]}
['funny', 'text', 'learning', 'learning', 'learning', 'learning', 'take', 'I', 'purpose', 'cruel']
预期的输出应为：（鉴于“学习”已经出现过一次，因此我们不应该再次包含它）
{0: [1, 0], 1: [0], 2: [1,0], 3: [1,2], 4: [0, 2, 8,4]} <---- (my idea is that since "learning" appears 3 times here, maybe we should take one time only despite which index it is in). With this, it could be 4:[0] or [2] or [8].
“ I”出现在2D列表的两个不同索引中。我的想法是采用任一索引，仅返回1次。可能是2：[1,0]或4：[0,4]或4：[2,4]或4：[8,4]
['funny', 'text', 'learning', 'take', 'I', 'purpose', 'cruel']
希望这也可以做到。我仍在尽力了解更多信息。谢谢您一直以来的帮助。
解决了！这里供以后参考：
store_x = {}
vex = []
for ex, x in enumerate(b):
row = []
for i, xx in enumerate(x):
if xx not in vex:
if xx in ba:
row.append(i)
vex.append(xx)
if row:
row.sort(key=lambda e: ba.index(x[e]))
store_x[ex] = row
print(store_x)
order = {e: ii for ii, e in enumerate(ba)}
dic2 = []
for i, x in store_x.items():
if x:
for ex in x:
dic2.append(a[i][ex])
dic2.sort(key=lambda e: order.get(e, len(ba)))
print(dic2)
答案 0 :(得分：1)
您可以这样做：
store_x = []
for x in b:
row = []
for i, xx in enumerate(x):
if xx in ba:
row.append(i)
store_x.append(row)
print(store_x)
dic2 = []
for i, x in enumerate(store_x):
row = []
if x:
for ex in x:
row.append(a[i][ex])
dic2.append(row)
print(dic2)
输出
[[0, 1], [], [], [2]]
[['text', 'man'], [], [], ['only']]
可以读取输出，因为store_x在索引0中具有值[0,1]，在索引3中具有值[2]。
您的代码中的问题是store_x是一个列表，您需要一个列表列表（二维列表）。一种选择是在第一个循环中使用字典：
store_x = {}
for ex, x in enumerate(b):
row = []
for i, xx in enumerate(x):
if xx in ba:
row.append(i)
if row:
store_x[ex] = row
print(store_x)
dic2 = []
for i, x in store_x.items():
row = []
if x:
for ex in x:
row.append(a[i][ex])
if row:
dic2.append(row)
print(dic2)
输出
{0: [0, 1], 3: [2]}
[['text', 'man'], ['only']]
这更像预期的输出。
更新 如注释中所指定的，假设ba缺少的值必须排在最后，则输出大多数遵循ba的顺序。该代码必须更新为：
store_x = {}
for ex, x in enumerate(b):
row = []
for i, xx in enumerate(x):
if xx in ba:
row.append(i)
if row:
store_x[ex] = row
print(store_x)
order = {e: ii for ii, e in enumerate(ba)}
dic2 = []
for i, x in store_x.items():
row = []
if x:
for ex in x:
row.append(a[i][ex])
if row:
row.sort(key=lambda e: order.get(e, len(ba)))
dic2.append(row)
print(dic2)
输出（测试用例1）
{0: [0, 1], 3: [2]}
[['text', 'man'], ['only']]
输出（测试用例2）
{0: [0], 2: [1], 3: [1, 2]}
[['text'], ['take'], ['purpose', 'learning']]
更新2
store_x = {}
for ex, x in enumerate(b):
row = []
for i, xx in enumerate(x):
if xx in ba:
row.append(i)
if row:
row.sort(key=lambda e: ba.index(x[e]))
store_x[ex] = row
print(store_x)
order = {e: ii for ii, e in enumerate(ba)}
dic2 = []
for i, x in store_x.items():
if x:
for ex in x:
dic2.append(a[i][ex])
dic2.sort(key=lambda e: order.get(e, len(ba)))
print(dic2)
输出（测试用例1）
{0: [0, 1], 3: [2]}
['text', 'man', 'only']
输出（测试用例2）
{0: [0], 2: [1], 3: [2, 1]}
['text', 'take', 'purpose', 'learning']
输出（测试用例3）
{0: [1, 0], 1: [0], 2: [1, 0], 3: [2, 1]}
['funny', 'text', 'take', 'I', 'purpose', 'learning', 'cruel']