Python Pandas-在具有dict列的数据框中添加/更新键值对

时间:2019-12-20 16:03:16

标签: python pandas dataframe dictionary key-value

具有此数据框

d = {'objects':[{'Sand':10},{'Seawater': 2, 'Crab': 30},{'Parasol': 50}]}
df = pd.DataFrame(data=d)

我想要这个键值对

{'Small': 1000}

插入具有至少小于40的键值对的每一行。

     objects
0    {'Sand': 10, 'Small': 1000}
1    {'Seawater': 2, 'Crab': 30, 'Small': 1000}
2    {'Parasol': 50}

我已经尝试过遍历它,但是会产生'None'。

def your_small(x):
    if any(value < 40 for value in x.values()):
        return x.update({'Small': 1000})

d = {'objects':[{'Sand':10},{'Seawater': 2, 'Crab': 30},{'Parasol': 50}]}
df = pd.DataFrame(data=d)
df['objects'] = df['objects'].map(your_small)
  objects
0    None
1    None
2    None

2 个答案:

答案 0 :(得分:3)

正如@ALoll所说,您可能要重新考虑您的方法。

如果要使现有代码正常工作,则必须考虑map的工作原理:必须在map函数中返回一个值。 x.update返回None,如果不满足条件,则必须按原样返回x:

def your_small(x):
    if any(value < 40 for value in x.values()):
        return {**x, 'Small': 1000}
    return x

答案 1 :(得分:2)

如果没有必要使用字典,则可以使用MultiIndex。在这里,我假设单独的字典主要具有不重叠的键,因此较长的DataFrame似乎更合适。 (如果大多数字典的键重叠,那么宽的DataFrame可能会更好)

import pandas as pd

df = pd.concat([pd.DataFrame.from_dict(di, orient='index', columns=['N']) for di in d['objects']], 
               keys=range(len(d['objects'])))
#             N
#0 Sand      10
#1 Seawater   2
#  Crab      30
#2 Parasol   50

# Determine which original "rows" have at least one value < 40
s = df.N.lt(40).groupby(level=0).any()

df_add = pd.DataFrame({'N': 1000},
                      index = pd.MultiIndex.from_product([s[s].index, ['Small']]))

# Join them:
df = pd.concat([df, df_add]).sort_index()
#               N
#0 Sand        10
#  Small     1000
#1 Crab        30
#  Seawater     2
#  Small     1000
#2 Parasol     50

这是一个具有广泛DataFrame的版本。易于操作,但会变得很大。

df = pd.DataFrame.from_records(d['objects'])
#   Sand  Seawater  Crab  Parasol
#0  10.0       NaN   NaN      NaN
#1   NaN       2.0  30.0      NaN
#2   NaN       NaN   NaN     50.0

df.loc[df.lt(40).any(1), 'Small'] = 1000
#   Sand  Seawater  Crab  Parasol   Small
#0  10.0       NaN   NaN      NaN  1000.0
#1   NaN       2.0  30.0      NaN  1000.0
#2   NaN       NaN   NaN     50.0     NaN