def check_baseline(a):
# 检查每个单词的baseline，如果baseline不同，则分离出去
# 字符的高度有两种，一等高度和二等高度
log.info("enter")
ans = []
for i in a:
baseline = []
for ch in i:
h = ch['max_x'] - ch['min_x']
if abs(h - 15) < 3: # 只统计占一格的字符
baseline.append(ch['max_x'])
if len(baseline) < 3:
ans.append(i)
continue
baseline_value = find_frequent(baseline)
new_word = []
for ch in i:
h = ch['max_x'] - ch['min_x']
if abs(h - 15) < 3: # 如果占一格
if abs(ch['max_x'] - baseline_value) > 4:
ans.append([ch]) # 占一格且远离基线，必然不属于当前word
else:
new_word.append(ch)
else:
new_word.append(ch)
ans.append(new_word)
log.info("end")
return ans
def go(img):
log.info("enter")
img, recs = img2recs(img=img)
if len(recs) == 0: return
recs = eye.predict(recs)
if len(recs) == 0: return
recs = sort_recs(recs)
ans = "".join(map(lambda i: i['ans'], recs))
print(imgId, ans)
log.info("end")
return ans
def grab_img():
# 获取屏幕截图
global imgId
log.info('enter')
# 图片越大截图花费时间越多，所以截图应该尽量小
# img = ImageGrab.grab((250, 161, 1141, 610))
filename = "grab/%d.jpg" % imgId
grab(filename, 62, 160, 857, 575)
imgId += 1
img = io.imread(filename)
log.info("end")
return img
def sort_words(words):
# 对单词进行排序，防止单词覆盖
# 优先拼写较长的单词，因为较长的单词表明该单词没有被覆盖
log.info("enter")
v = [0] * len(words)
for ind, i in enumerate(words):
w = ''.join(map(lambda x: x['ans'], i))
print(w)
if w in english_dic:
v[ind] = len(w)
v = np.argsort(v)
a = []
for i in range(len(v) - 1, -1, -1):
a.append(words[v[i]])
log.info("end")
return a
def build_words(recs):
# 根据小矩形构建单词
log.info("enter")
char_gap_width = 9
recs = sorted(recs, key=lambda x: x['min_y'])
a = []
for i in recs:
had = False
for ind, word in enumerate(a):
if word[-1]['max_y'] + char_gap_width > i['min_y'] and intersect(
word[-1], i):
word.append(i)
had = True
if not had:
a.append([i])
log.info("end")
return a
def find_frequent(a):
"""
给定一个数组，求数组中的众数
方法是对元素进行KMEANS聚类，聚成3类，取元素最多的那一类
这种方法可以寻找近似众数
:param a:
:return:
"""
log.info("enter")
a = np.array(a)
km = KMeans(n_clusters=min(len(a), 3))
label = km.fit_predict(np.reshape(a, (-1, 1)))
cnt = collections.Counter(label)
ma = None
for i in cnt:
if ma is None or cnt[ma] < cnt[i]:
ma = i
baseline = np.mean(a[label == ma])
log.info("end")
return baseline