Question

我正在使用Pandas的DataFrames。合并2个文件后，我遇到这样的情况： fr.item = "ipod"; fr.bucket = {'ipad':34,'ipod':36,'iwatch':27} 注意：数据类型为Series

是否可以在此处检查存储桶（ipod）中的项目并获取值（36）？另外，不加循环也很感激，因为我正在进行列与列的比较。

输入

item    bucket
ipod    {'ipad':34,'ipod':36,'iwatch':27}
ipad    {'ipad':87,'ipod':31,'iwatch':62}

输出

36
87

Answer 1

同意@bazingaa，只需使用：

fr.bucket[fr.item]

因此，使用命名为ipod的字典键，然后像上面一样获取它的值。

Answer 2

尝试一下：

# update col1 and col2 to the name of your item and bucket column
col1 = 'item'
col2 = 'bucket'
output = fr.apply(lambda row: row[col2][col1], axis=1)

它将返回熊猫系列36、87。

更新：如果您在存储桶列中的商品的类型为str，请首先尝试以下操作：

import ast
fr.loc[:, col2] = fr[col2].apply(ast.literal_eval)

Answer 3

item=['ipod','ipad']
bucket=[{'ipad':34,'ipod':36,'iwatch':27},{'ipad':87,'ipod':31,'iwatch':62}]

result_output=[]    # Defining for output
counter=0           # For initialising the counter through the bucket's elements
for key in item:    # For each row value in column1. Column1 is item.
   temp_bucket=dict(bucket[counter]) # For each column's value,it is as a dictionary
   if key in temp_bucket:   # Checking if the column1's value is present in the dictionary of that row (column2)   
      length_1=len(bucket)-1                 # Need to subtract 1 as python is based on 0 indexing     
      result_output.append(temp_bucket[key]) # Appending each row output to the result list.
      if counter<length_1:                   # Need to check if there is sufficient element is present in column2. Otherwise throw error for index.
         counter=counter+1                  # for checking the next element of column2

print（result_output）＃输出列表

Answer 4

我尝试过，它有效

对于df.index中的i：

#For Bucket
val_bucket=ast.literal_eval(bucket[i])
for key, val in val_bucket.items():
    if my_items[i]== key:
        print ("Value",val)
         break

因此，我已经将这些JSON值（串联）转换为dict并提取了值

使用Pandas将字符串转换为JSON比较

4 个答案: