我是python新手，我為此苦苦掙扎了一段時間。我有一個看起來像這樣的文件： name seq1 a1 bbb2 a2 bbc3 b1 fff4 b2 fff5 c1 aaa6 c2 acg其中 name 是字符串的名稱，seq 是字符串。我想要一個新列或一個新數據框來指示每兩行之間沒有重疊的差異數量。例如，我想要名稱 [a1-a2] 然后 [b1-b2] 和最后 [c1-c2] 之間的序列之間的差異數。所以我需要這樣的東西： name seq diff 1 a1 bbb NA 2 a2 bbc 13 b1 fff NA4 b2 fff 05 c1 aaa NA6 c2 acg 2非常感謝任何幫助
4 回答
慕沐林林
TA貢獻2016條經驗 獲得超9個贊
看起來您想要字符串對的杰卡德距離。groupby這是使用and的一種方法scipy.spatial.distance.jaccard：
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
飲歌長嘯
TA貢獻1951條經驗 獲得超3個贊
Levenshtein距離替代：
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
鴻蒙傳說
TA貢獻1865條經驗 獲得超7個贊
作為第一步，我使用以下方法重新創建了您的數據：
#!/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)
解決方案 您可以嘗試迭代數據框并將seq最后一次迭代的值與當前迭代值進行比較。為了比較兩個字符串（存儲在數據框的seq列中），您可以應用一個簡單的列表推導，如在此函數中：
def diff_letters(a,b):
return sum ( a[i] != b[i] for i in range(len(a)) )
迭代 Dataframe 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
結果看起來像這樣
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
侃侃爾雅
TA貢獻1801條經驗 獲得超16個贊
檢查這個
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
結果是這樣的：
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
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