Question

我有一个数据帧df，如下所示：

        a      b
  0     Jon     Jon
  1     Jon     John
  2     Jon     Johnny

我想将这两个字符串进行比较，并新建一个这样的列：

  df['compare'] = df2['a'] = df2['b']


        a      b          compare
  0     Jon     Jon         True
  1     Jon     John        False
  2     Jon     Johnny      False

我还希望能够通过levenshtein函数传递列a和b：

def levenshtein_distance(a, b):
    """Return the Levenshtein edit distance between two strings *a* and *b*."""
    if a == b:
        return 0
    if len(a) < len(b):
        a, b = b, a
    if not a:
        return len(b)
    previous_row = range(len(b) + 1)
    for i, column1 in enumerate(a):
        current_row = [i + 1]
        for j, column2 in enumerate(b):
            insertions = previous_row[j + 1] + 1
            deletions = current_row[j] + 1
            substitutions = previous_row[j] + (column1 != column2)
            current_row.append(min(insertions, deletions, substitutions))
        previous_row = current_row
    return previous_row[-1]

并添加如下所示的列：

  df['compare'] = levenshtein_distance(df2['a'], df2['b'])      

        a      b          compare
   0    Jon     Jon         100
   1    Jon     John        .95
   2    Jon     Johnny      .87

但是尝试时出现此错误：

  ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

如何格式化我的数据/数据框以允许它比较两列，并将比较添加到第三列？

Answer 1

只需：

df['compare'] = [levenshtein_distance(a, b) for a, b in zip(df2['a'], df2['b'])]

或者，如果要进行相等比较：

df['compare'] = (df['a'] == df['b'])

Answer 2

我认为您比较是错误的，请更改：

更改：

if a == b

and not a

到

if a[0] == b[0]

and 

not a[0]

，您将看到您的函数有效，它只需要遍历所传递的df。如果返回列表，则等于将返回

这是一个有效的版本：

def levenshtein_distance(a, b):
  """Return the Levenshtein edit distance between two strings *a* and *b*."""
  y = len(a)
  thelist = []
  for x in range(0, y):
    c = a[x]
    d = b[x] 
    if c == d:
        thelist.append(0)
        continue
    if len(c) < len(d):
        c, d = d, c
    if not c:
        thelist.append(len(d))
        continue
    previous_row = range(len(d) + 1)
    for i, column1 in enumerate(c):
        current_row = [i + 1]
        for j, column2 in enumerate(d):
            insertions = previous_row[j + 1] + 1
            deletions = current_row[j] + 1
            substitutions = previous_row[j] + (column1 != column2)
            current_row.append(min(insertions, deletions, substitutions))
        previous_row = current_row
    thelist.append(previous_row[-1])
  return thelist

df['compare'] =  levenshtein_distance(df.a, df.b)                                                                                                                

df                                                                                                                                                               

#     a       b  compare
#0  Jon     Jon        0
#1  Jon    John        1
#2  Jon  Johnny        3

它只是不计算百分比，仅使用您的代码，根据Levenshtein Calc，这是正确的答案

如何字符串比较熊猫数据框中的两列？

2 个答案: