Question

我有一个单词列表，我想测试字谜。我想使用熊猫，所以我不必使用计算浪费的循环。给定.txt单词列表说：

“ACB” “BCA” “富” “钱币” “犬”

我想将它们放在df中，然后按照它们的字谜列表对它们进行分组 - 我可以稍后删除重复的行。

到目前为止，我有代码：

import pandas as pd

wordlist = pd.read_csv('data/example.txt', sep='\r', header=None, index_col=None, names=['word'])
wordlist = wordlist.drop_duplicates(keep='first')
wordlist['split'] = ''
wordlist['anagrams'] = ''

for index, row in wordlist.iterrows() :
    row['split'] = list(row['word'])

wordlist = wordlist.groupby('word')[('split')].apply(list)
print(wordlist)

如何分组，以便知道

[[a, b, c]]
[[b, a, c]]

是一样的吗？

Answer 1

我认为您可以使用sorted list s：

df['a'] = df['word'].apply(lambda x: sorted(list(x)))
print (df)

      word                      a
0      acb              [a, b, c]
1      bca              [a, b, c]
2      foo              [f, o, o]
3      oof              [f, o, o]
4  spaniel  [a, e, i, l, n, p, s]

查找字谜的另一种解决方案：

#reverse strings
df['reversed'] = df['word'].str[::-1]

#reshape
s = df.stack()
#get all dupes - anagrams
s1 = s[s.duplicated(keep=False)]
print (s1)
0  word        acb
   reversed    bca
1  word        bca
   reversed    acb
2  word        foo
   reversed    oof
3  word        oof
   reversed    foo
dtype: object

#if want select of values by second level word
s2 = s1.loc[pd.IndexSlice[:, 'word']]
print (s2)
0    acb
1    bca
2    foo
3    oof
dtype: object

Df groupby设置比较

1 个答案: