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如何为pandas构造嵌套for循环和条件的列表理解?

list-comprehension list pandas python-3.x python

ShanZhengYang · 技术社区 · 6 年前

我很难像预期的那样理解下面的复杂列表。它是一个带条件的双嵌套for循环。

import pandas as pd

dict1 = {'stringA':['ABCDBAABDCBD','BBXB'], 'stringB':['ABDCXXXBDDDD', 'AAAB'], 'num':[42, 13]}

df = pd.DataFrame(dict1)
print(df)
        stringA       stringB  num
0  ABCDBAABDCBD  ABDCXXXBDDDD   42
1          BBXB          AAAB   13

此数据帧有两列 stringA 和 stringB 包含字符的字符串 A , B , C D , X . 根据定义,这两个字符串具有相同的长度。

基于这两列,我创建了这样的字典 斯特林加 字符串B 从开始于的索引开始 num .

def create_translation(x):
    x['translated_dictionary'] = {i: i +x['num'] for i, e in enumerate(x['stringA'])}
    return x

df2 = df.apply(create_translation, axis=1).groupby('stringA')['translated_dictionary']


df2.head()
0    {0: 42, 1: 43, 2: 44, 3: 45, 4: 46, 5: 47, 6: ...
1                         {0: 13, 1: 14, 2: 15, 3: 16}
Name: translated_dictionary, dtype: object

print(df2.head()[0])
{0: 42, 1: 43, 2: 44, 3: 45, 4: 46, 5: 47, 6: 48, 7: 49, 8: 50, 9: 51, 10: 52, 11: 53}

print(df2.head()[1])
{0: 13, 1: 14, 2: 15, 3: 16}

没错。

但是,这些字符串中有“X”字符。这需要一个特殊的规则:如果 十 在 斯特林加 ,不要在字典中创建键值对。如果在 字符串B ,则该值不应为 i + x['num'] 但是 -500 .

我尝试了以下列表:

def try1(x):
    for count, element in enumerate(x['stringB']):
        x['translated_dictionary'] = {i: -500 if element == 'X' else  i + x['num'] for i, e in enumerate(x['stringA']) if e != 'X'}
    return x

这是错误的答案。

df3 = df.apply(try1, axis=1).groupby('stringA')['translated_dictionary']

print(df3.head()[0]) ## this is wrong!
{0: 42, 1: 43, 2: 44, 3: 45, 4: 46, 5: 47, 6: 48, 7: 49, 8: 50, 9: 51, 10: 52, 11: 53}

print(df3.head()[1])   ## this is correct! There is no key for 2:15!
{0: 13, 1: 14, 3: 16}

正确答案是:

print(df3.head()[0])
{0: 42, 1: 43, 2: 44, 3: 45, 4:-500, 5:-500, 6:-500, 7: 49, 8: 50, 9: 51, 10: 52, 11: 53}

print(df3.head()[1])
{0: 13, 1: 14, 3: 16}

2 回复 | 直到 6 年前

John Zwinck 6 年前

这里有一个简单的方法,没有任何理解(因为它们无助于澄清代码):

def create_translation(x):
    out = {}
    num = x['num']
    for i, (a, b) in enumerate(zip(x['stringA'], x['stringB'])):
        if a == 'X':
            pass
        elif b == 'X':
            out[i] = -500
        else:
            out[i] = num
        num += 1
    x['translated_dictionary'] = out
    return x

BENY 6 年前

post 重新创造 dict

n=df.stringA.str.len()
newdf=pd.DataFrame({'num':df.num.repeat(n),'stringA':sum(list(map(list,df.stringA)),[]),'stringB':sum(list(map(list,df.stringB)),[])})


newdf=newdf.loc[newdf.stringA!='X'].copy()# remove stringA value X
newdf['value']=newdf.groupby('num').cumcount()+newdf.num # using groupby create the cumcount 
newdf.loc[newdf.stringB=='X','value']=-500# assign -500 when stringB is X
[dict(zip(x.groupby('num').cumcount(),x['value']))for _,x in newdf.groupby('num')] # create the dict for different num by group
Out[390]: 
[{0: 13, 1: 14, 2: 15},
 {0: 42,
  1: 43,
  2: 44,
  3: 45,
  4: -500,
  5: -500,
  6: -500,
  7: 49,
  8: 50,
  9: 51,
  10: 52,
  11: 53}]