检查2D列表的索引并根据2D列表打印出索引

Question

我想尝试另一个示例，以打印2D列表的索引。例子：

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)

电流输出：

[0, 1, 2]

['text', 'just', 'this']

预期产量：

[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，然后从索引值中，使用存储在{store_x}中的索引，使用它从{a}打印值。

Answer 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']]输出可以读取为store_x索引 0 中的值[0,1]和索引 3 中的值[2]。您的代码中的问题是这store_x是一个列表，您需要一个列表列表（2D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] = rowprint(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']]这更像是预期的输出。UPDATE 如注释中所指定，输出最遵循ba的顺序，假设ba缺少的值必须排在最后。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] = rowprint(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2store_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] = rowprint(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']