查找每两行 pandas data.frame 的字符串之间的差异

Question

我是 python 新手，我为此苦苦挣扎了一段时间。 我有一个看起来像这样的文件：

    name   seq
1   a1     bbb
2   a2     bbc
3   b1     fff
4   b2     fff
5   c1     aaa
6   c2     acg

其中 name 是字符串的名称，seq 是字符串。 我想要一个新列或一个新数据框，指示每两行之间没有重叠的差异数量。例如，我想要名称 [a1-a2] 然后 [b1-b2] 和最后 [c1-c2] 之间的序列之间的差异数。

所以我需要这样的东西：

    name   seq   diff  
1   a1     bbb    NA   
2   a2     bbc    1
3   b1     fff    NA
4   b2     fff    0
5   c1     aaa    NA
6   c2     acg    2

非常感谢任何帮助

Answer 1

看起来您想要字符串对中的jaccard distance。这是使用groupby 和scipy.spatial.distance.jaccard 的一种方法：

from scipy.spatial.distance import jaccard
g = df.groupby(df.name.str[0])

df['diff'] = [sim for _, seqs in g.seq for sim in 
              [float('nan'), jaccard(*map(list,seqs))]]

---

print(df)

  name  seq  diff
1   a1  bbb   NaN
2   a2  bbc   1.0
3   b1  fff   NaN
4   b2  fff   0.0
5   c1  aaa   NaN
6   c2  acg   2.0

Answer 2

替代Levenshtein距离：

import Levenshtein
s = df['name'].str[0]
out = df.assign(Diff=s.drop_duplicates(keep='last').map(df.groupby(s)['seq']
                    .apply(lambda x: Levenshtein.distance(x.iloc[0],x.iloc[-1]))))

---

  name  seq  Diff
1   a1  bbb   NaN
2   a2  bbc   1.0
3   b1  fff   NaN
4   b2  fff   0.0
5   c1  aaa   NaN
6   c2  acg   2.0

Answer 3

作为第一步，我使用以下方法重新创建了您的数据：

#!/usr/bin/env python3
import pandas as pd

# Setup
data = {'name': {1: 'a1', 2: 'a2', 3: 'b1', 4: 'b2', 5: 'c1', 6: 'c2'}, 'seq': {1: 'bbb', 2: 'bbc', 3: 'fff', 4: 'fff', 5: 'aaa', 6: 'acg'}}
df = pd.DataFrame(data)

解决方案 您可以尝试迭代数据框并将最后一次迭代的seq 值与当前值进行比较。为了比较两个字符串（存储在数据框的 seq 列中），您可以应用一个简单的列表理解，例如在此函数中：

def diff_letters(a,b):
    return sum ( a[i] != b[i] for i in range(len(a)) )

迭代数据框行

diff = ['NA']

row_iterator = df.iterrows()
_, last = next(row_iterator)

# Iterate over the df get populate a list with result of the comparison
for i, row in row_iterator:
    if i % 2 == 0:
        diff.append(diff_letters(last['seq'],row['seq']))
    else:
        # for odd row numbers append NA value
        diff.append("NA")
    last = row
df['diff'] = diff

结果如下所示

  name  seq diff
1   a1  bbb   NA
2   a2  bbc    1
3   b1  fff   NA
4   b2  fff    0
5   c1  aaa   NA
6   c2  acg    2

Answer 4

检查这个

import pandas as pd

data = {'name':  ['a1', 'a2','b1','b2','c1','c2'],
    'seq': ['bbb', 'bbc','fff','fff','aaa','acg']
    }

df = pd.DataFrame (data, columns = ['name','seq'])
diffCntr=0
df['diff'] = np.nan
i=0
while i < len(df)-1:
    diffCntr=np.nan
    item=df.at[i,'seq']
    df.at[i,'diff']=diffCntr
    diffCntr=0
    for j in df.at[i+1,'seq']:
        if item.find(j) < 0:
            diffCntr +=1
    df.at[i+1,'diff']=diffCntr
    i +=2    
df  

结果是这样的：

    name seq    diff
0   a1   bbb    NaN
1   a2   bbc    1.0
2   b1   fff    NaN
3   b2   fff    0.0
4   c1   aaa    NaN
5   c2   acg    2.0