如何在 Python 中查找列表中的字符串之间的相似性

Question

我正在比较 Python 中的两个数据框列，目的是为第一列的每个元素找到第二列的最佳匹配。第一列包含 19.000 行，我需要检查它的每个字符串，第二列的最佳匹配是什么。因此，需要检查 19.000 行，每行 19.000 次，考虑到字符串本身必须是另一个，而不是相同的。

我从一个简单的比较开始，在列表中找到一个字符串，我成功了。然后我将它应用于一个列表，只是为了比较它们，但显然，由于比较字符串与列表，会给出错误“TypeError：预期的字符串或类似字节的对象”。最后，我尝试创建一个循环，但错误是一样的。有没有办法创建一个具有预期结果的列表？也许有更好的方法来使用另一个库，但是，到目前为止，我什么也没找到。这是目前的代码：

#simple example
from fuzzywuzzy import process
string = "appl"
compare = ["adfad.","apple","asple","tab"]
Ratios = process.extract(string,compare)
print(Ratios)
[('apple', 89), ('asple', 67), ('tab', 29), ('adfad.', 22)]

highest = process.extractOne(string,compare)
print(highest)
('apple', 89)

#data frame
from fuzzywuzzy import process
dataframecolumn = ["appl","tb"]
compare = ["adfad.","apple","asple","tab"]
Ratios = process.extract(dataframecolumn,compare)
TypeError: expected string or bytes-like object

#expected (but I need a list)
highest = process.extractOne(dataframecolumn[0],compare)
print(highest)
('apple', 89)
highest = process.extractOne(dataframecolumn[1],compare)
print(highest)
('tab', 80)

#Result expected
results = ["apple, 89","tab, 80"]

#Error
myl = ["appl","tb"]
compare = ["adfad.","apple","asple","tab"]
results = []
for x in myl:
    results.append(process.extractOne(myl,compare)[1])
TypeError: expected string or bytes-like object

Answer 1

from operator import itemgetter 

dataframecolumn = ["appl","tb"]
compare = ["adfad.","apple","asple","tab"]
Ratios = [process.extract(x,compare) for x in dataframecolumn]
print ([max(ratios, key = itemgetter(1)) for ratios in Ratios])

# Or oneliner
#Ratios = [max(process.extract(x,compare),key = itemgetter(1)) for x in dataframecolumn]

如果extract 总是返回排序后的结果，那么我们可以避免调用max

Ratios = [process.extract(x, compare)[0] for x in dataframecolumn]

输出：

[('apple', 89), ('tab', 80)]

如果您想跳过精确匹配而只获得模糊匹配，则只需跳过得分为 100% 的匹配并获得第一个非 100% 匹配，因为它已经排序。

dataframecolumn = ["apple","tb"]
compare = ["adfad","apple","asple","tab"]
Ratios = [process.extract(x,compare) for x in dataframecolumn]
result = list()
for ratio in Ratios:
    for match in ratio:
        if match[1] != 100:
            result.append(match)
            break
print (result)