Python 如果存在差异，是否获取具有相同id的项目的最小最大值？
我有两个看起来像这样的文件，它们之间有些不同： 第一个文件：Python 如果存在差异，是否获取具有相同id的项目的最小最大值？,python,difflib,Python,Difflib,我有两个看起来像这样的文件，它们之间有些不同： 第一个文件： {16:[3, [-7, 87, 20, 32]]} {17:[2, [-3, 88, 16, 28], 3, [-6, 84, 20, 32]]} {18:[2, [-1, 88, 16, 28], 3, [-3, 84, 20, 32]]} {19:[2, [1, 89, 16, 28], 3, [-2, 85, 20, 32]]} {20:[2, [9, 94, 16, 28], 3, [1, 85, 20, 32]]} {2
{16:[3, [-7, 87, 20, 32]]}
{17:[2, [-3, 88, 16, 28], 3, [-6, 84, 20, 32]]}
{18:[2, [-1, 88, 16, 28], 3, [-3, 84, 20, 32]]}
{19:[2, [1, 89, 16, 28], 3, [-2, 85, 20, 32]]}
{20:[2, [9, 94, 16, 28], 3, [1, 85, 20, 32]]}
{21:[2, [12, 96, 16, 28], 3, [2, 76, 19, 31]]}
{22:[2, [15, 97, 16, 28], 3, [4, 73, 19, 29]]}
{23:[2, [18, 96, 16, 28], 3, [6, 71, 19, 29], 10, [-10, 60, 51, 82]]}
{24:[2, [22, 97, 16, 28], 3, [9, 71, 19, 27], 10, [-5, 63, 49, 78]]}
{25:[2, [25, 99, 16, 28], 3, [13, 71, 17, 26], 10, [-1, 64, 46, 77]]}
{26:[2, [29, 101, 16, 28], 3, [17, 70, 16, 25], 10, [-1, 65, 45, 77]]}
{16:[3, [-7, 86, 20, 32]]}
{17:[2, [-3, 82, 16, 28], 3, [-6, 84, 20, 32]]}
{18:[2, [-1, 88, 16, 27], 3, [-3, 84, 20, 32]]}
{19:[2, [1, 89, 16, 28], 3, [-2, 84, 20, 32]]}
{20:[2, [9, 94, 15, 28], 3, [1, 85, 20, 32]]}
{21:[2, [12, 96, 16, 28], 3, [1, 76, 19, 31]]}
{22:[2, [15, 97, 17, 28], 3, [4, 73, 19, 29]]}
{23:[2, [18, 96, 18, 28], 3, [6, 71, 19, 29], 10, [-10, 60, 51, 82]]}
{24:[2, [22, 97, 16, 28], 3, [9, 71, 20, 27], 10, [-5, 63, 49, 78]]}
{25:[2, [25, 99, 16, 28], 3, [13, 71, 17, 26], 10, [-1, 64, 46, 77]]}
{26:[2, [29, 101, 17, 28], 3, [17, 70, 16, 25], 10, [-1, 65, 45, 77]]}
id帧162632317 - 36
3帧之一与我正在比较的文件不同
对于这样的每个差异，我需要写出一个只包含开始帧和结束帧的新文件，然后我将致力于将附加字符串连接到每个文件
这是当前difflib使用情况，它打印出具有不同
def compare(f1, f2):
with open(f1+'.txt', 'r') as fin1, open(f2+'.txt', 'r') as fin2:
diff = difflib.ndiff(fin1.readlines(), fin2.readlines())
outcome = ''.join(x[2:] for x in diff if x.startswith('- '))
print outcome
通过调整此执行块，我将如何实现上面所述的功能
请注意，这两个文件共享相同的frame
amount，但不相同的id
s，因此我需要为每个差异编写两个不同的文件，可能会写入一个文件夹。因此，如果这两个文件有20个差异，我需要有两个主文件夹，每个原始文件一个，每个原始文件包含相同id的每个开始和结束帧的文本文件。假设您的差异列表是您在文章开头给出的文件内容。我进行了2次，第一次获得每个id的帧列表：
>>> from collections import defaultdict
>>> diffs = defaultdict(list)
>>> for line in s.split('\n'):
d = eval(line) # We have a dict
for k in d: # Only one value, k is the frame
# Only get even values for ids
for i in range(0, len(d[k]), 2):
diffs[d[k][i]].append(k)
>>> diffs # We now have a dict with ids as keys :
defaultdict(<type 'list'>, {10: [23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36], 2: [17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33], 3: [16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36], 29: [31, 32, 33, 34, 35, 36]})
然后，对于每个id，都有差异的范围。我想只要稍加调整，你就能得到你想要的
编辑：以下是相同类型块的最终答案：
>>> def compare(f1, f2):
# 2 embedded 'with' because I'm on Python 2.5 :-)
with open(f1+'.txt', 'r') as fin1:
with open(f2+'.txt', 'r') as fin2:
lines1 = fin1.readlines()
lines2 = fin2.readlines()
# Do not forget the strip function to remove unnecessary '\n'
diff_lines = [l.strip() for l in lines1 if l not in lines2]
# Ok, we have our differences (very basic)
diffs = defaultdict(list)
for line in diff_lines:
d = eval(line) # We have a dict
for k in d:
list_ids = d[k] # Only one value, k is the frame
for i in range(0, len(d[k]), 2):
diffs[d[k][i]].append(k)
for id_ in diffs:
diffs[id_].sort()
for k, g in groupby(enumerate(diffs[id_]), lambda (i, x): i - x):
group = map(itemgetter(1), g)
print 'id {0} : {1} -> {2}'.format(id_, group[0], group[-1])
>>> compare(r'E:\CFM\Dev\Python\test\f1', r'E:\CFM\Dev\Python\test\f2')
id 2 : 17 -> 24
id 2 : 26 -> 26
id 3 : 16 -> 24
id 3 : 26 -> 26
id 10 : 23 -> 24
id 10 : 26 -> 26
假设您的差异列表是您在文章开头给出的文件内容。我进行了2次，第一次获得每个id的帧列表：
>>> from collections import defaultdict
>>> diffs = defaultdict(list)
>>> for line in s.split('\n'):
d = eval(line) # We have a dict
for k in d: # Only one value, k is the frame
# Only get even values for ids
for i in range(0, len(d[k]), 2):
diffs[d[k][i]].append(k)
>>> diffs # We now have a dict with ids as keys :
defaultdict(<type 'list'>, {10: [23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36], 2: [17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33], 3: [16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36], 29: [31, 32, 33, 34, 35, 36]})
然后，对于每个id，都有差异的范围。我想只要稍加调整，你就能得到你想要的
编辑：以下是相同类型块的最终答案：
>>> def compare(f1, f2):
# 2 embedded 'with' because I'm on Python 2.5 :-)
with open(f1+'.txt', 'r') as fin1:
with open(f2+'.txt', 'r') as fin2:
lines1 = fin1.readlines()
lines2 = fin2.readlines()
# Do not forget the strip function to remove unnecessary '\n'
diff_lines = [l.strip() for l in lines1 if l not in lines2]
# Ok, we have our differences (very basic)
diffs = defaultdict(list)
for line in diff_lines:
d = eval(line) # We have a dict
for k in d:
list_ids = d[k] # Only one value, k is the frame
for i in range(0, len(d[k]), 2):
diffs[d[k][i]].append(k)
for id_ in diffs:
diffs[id_].sort()
for k, g in groupby(enumerate(diffs[id_]), lambda (i, x): i - x):
group = map(itemgetter(1), g)
print 'id {0} : {1} -> {2}'.format(id_, group[0], group[-1])
>>> compare(r'E:\CFM\Dev\Python\test\f1', r'E:\CFM\Dev\Python\test\f2')
id 2 : 17 -> 24
id 2 : 26 -> 26
id 3 : 16 -> 24
id 3 : 26 -> 26
id 10 : 23 -> 24
id 10 : 26 -> 26
这很有帮助，非常感谢！但是，我不明白s.split（'\n'）中第三行中的“s”代表什么：哦，是的，对不起：-）s
只是一个大字符串，包含了你在开头显示的所有行，它代表了差异列表。唯一的问题是我需要它来打印每个id的最小-最大帧，所以id 2:17->26左右…但是再次感谢，我会用它来解决问题。哦，好的，我想你你想要所有的范围！你只需要将diff中id的后面的所有复杂逻辑替换为：print min（diff[id]），max（diff[id]）
就可以了。这非常有用，非常感谢！但是，我不明白s.split（'\n'）中第三行的“s”是什么意思：哦，是的，抱歉：-）s
只是一个大字符串，包含开头显示的所有行，它代表差异列表。唯一的问题是我需要它打印每个id的最小-最大帧，所以id 2:17->26左右。。。但再次感谢你，我会解决这个问题，并尝试找出解决方案。哦，好的，我还以为你想要所有的范围呢！您只需将diff:
中id的之后的所有复杂逻辑替换为：printmin（diff[id]）、max（diff[id]）
，就可以完成了。