Question

我有DataFrame：

import pandas as pd
df = pd.DataFrame({'Board': ['A', 'B'], 'Off': ['C', 'D'], 'Stops': ['Q/W/E', 'Z'], 'Pax': [10, 100]})

看起来像：

    Board   Off  Pax    Stops
0   A       C    10     Q/W/E
1   B       D    100    Z

我希望按DataFrame列进行Stops拆分，并将Board和Off重新排列为Pax值重复的行，如下所示;

    Board   Off  Pax
0   A       Q    10
1   Q       W    10
2   W       E    10
3   E       C    10
4   B       Z    100
5   Z       D    100

对此有任何帮助将不胜感激。

Answer 1

我打破了步骤

df['New']=df[['Board','Stops','Off']].apply(lambda x : '/'.join(x),1)
df['New2']=df['New'].str.split('/').apply(lambda x : list(zip(x[:-1],x[1:])))

namedict = {0 : 'Board',1:'Off'}
df[['Pax','New2']].set_index('Pax').New2.apply(pd.Series).\ 
       stack().apply(pd.Series).reset_index().\
             drop('level_1',1).rename(columns=namedict)
Out[1260]: 
   Pax Board Off
0   10     A   Q
1   10     Q   W
2   10     W   E
3   10     E   C
4  100     B   Z
5  100     Z   D

Answer 2

以下是使用拆分和pd.concat的一种方法，使用时会将元组转换为sereis pd.DataFrame

#Split the  by `/`
df['New'] = (df['Board']+'/'+df['Stops']+'/'+df['Off']).str.split('/')
#0    [A, Q, W, E, C]
#1          [B, Z, D]
#Name: New, dtype: object

temp = df['New'].apply(lambda x : list(zip(x,x[1:])))

# Zip will lead to output given (1,2,3) -> [(1,2),(2,3)]
#0    [(A, Q), (Q, W), (W, E), (E, C)]
#1                    [(B, Z), (Z, D)]
#Name: New, dtype: object

di = {0 : 'Board',1:'Off','index':'Pax'}
# Concat each row by converting it to DataFrame with key as Pax and rename the columns. 

ndf = pd.concat([pd.DataFrame(i,index=np.repeat(j,len(i))) for (i,j) in zip(temp,df['Pax'].values)])\
     .reset_index().rename(columns=di)

输出：

  Pax Board Off
0   10     A   Q
1   10     Q   W
2   10     W   E
3   10     E   C
0  100     B   Z
1  100     Z   D

Answer 3

from itertools import islice
#https://stackoverflow.com/a/6822773/2901002
#added a for return Pax
def window(a, seq, n=2):
    "Returns a sliding window (of width n) over data from the iterable"
    "   s -> (s0,s1,...s[n-1]), (s1,s2,...,sn), ...                   "
    it = iter(seq)
    result = tuple(islice(it, n))
    if len(result) == n:
        yield (*result, a)
    for elem in it:
        result = result[1:] + (elem,)
        yield (*result, a)

#join all columns 
a = df['Board'] + '/' + df['Stops'] + '/' + df['Off']
#split to lists
a = a.str.split('/')

#apply window function to each value and flatten
L = [(j,k,l) for x, y in zip(a, df['Pax']) for j,k,l in list(window(y, x, 2)) ]
print (L)
[('A', 'Q', 10), ('Q', 'W', 10), ('W', 'E', 10), 
 ('E', 'C', 10), ('B', 'Z', 100), ('Z', 'D', 100)]

#DataFrame constructor
df = pd.DataFrame(L, columns=['Board','Off','Pax'])
print (df)
  Board Off  Pax
0     A   Q   10
1     Q   W   10
2     W   E   10
3     E   C   10
4     B   Z  100
5     Z   D  100

<强>计时：

import pandas as pd
N = 1000
L1 = list('ABCDEFGHIJKLMNOP')
L = ['Q/W/E','Q1/W1/E1','Z','A/B/C/D']
df = pd.DataFrame({'Board': np.random.choice(L1, N), 
                    'Off': np.random.choice(L1, N), 
                    'Stops': np.random.choice(L, N), 
                    'Pax': np.random.randint(100, size=N)})
print (df)

def bharath(df):
    df['New'] = (df['Board']+'/'+df['Stops']+'/'+df['Off']).str.split('/')
    temp = df['New'].apply(lambda x : list(zip(x,x[1:])))
    di = {0 : 'Board',1:'Off'}
    return pd.concat([pd.DataFrame(i,index=np.repeat(j,len(i))) for (i,j) in zip(temp,df['Pax'].values)]).reset_index().rename(columns=di)


def jez(df):
    from itertools import islice
    #https://stackoverflow.com/a/6822773/2901002
    #added a for return Pax
    def window(a, seq, n=2):
        "Returns a sliding window (of width n) over data from the iterable"
        "   s -> (s0,s1,...s[n-1]), (s1,s2,...,sn), ...                   "
        it = iter(seq)
        result = tuple(islice(it, n))
        if len(result) == n:
            yield (*result, a)
        for elem in it:
            result = result[1:] + (elem,)
            yield (*result, a)

    a = df['Board'] + '/' + df['Stops'] + '/' + df['Off']
    a = a.str.split('/')
    L = [(j,k,l) for x, y in zip(a, df['Pax']) for j,k,l in list(window(y, x, 2)) ]
    return pd.DataFrame(L, columns=['Board','Off','Pax'])

def wen(df):
    df['New']=df[['Board','Stops','Off']].apply(lambda x : '/'.join(x),1)
    df['New2']= df['New'].str.split('/').apply(lambda x : list(zip(x[:-1],x[1:])))
    namedict = {0 : 'Board',1:'Off'}
    return df[['Pax','New2']].set_index('Pax').New2.apply(pd.Series).stack().apply(pd.Series).reset_index().drop('level_1',1).rename(columns=namedict)



print (jez(df))
#print (bharath(df))
print (wen(df))

In [433]: %timeit (jez(df))
100 loops, best of 3: 6.6 ms per loop

In [434]: %timeit (wen(df))
1 loop, best of 3: 747 ms per loop

In [450]: %timeit (bharath(df))
1 loop, best of 3: 406 ms per loop

Answer 4

相对更具可读性和效率：

def expand_rows(series):
    """
    expand a row by splitting on "/"
    """
    return series.str.split("/", expand=True).stack()

d = {'Board': lambda dff: dff['Board'] + '/' + dff['Stops'],
     'Off': lambda dff: dff['Stops'] + '/' + dff['Off']
    }

df.set_index('Pax').assign(**d)[['Board', 'Off']].apply(expand_rows).reset_index(level=0)

使用来自@ jezrael帖子的数据集计时，它产生以下结果：

每回路15.8 ms±146μs（平均值±标准偏差，7次运行，每次100次循环）

Pandas：拆分数据帧行并重新排列列值

4 个答案: