为什么我们需要将压缩的对象转换为列表 [英] Why do we need to convert a zipped object into a list

问题描述

我正在尝试完成一个数据营练习，其中需要将2个列表转换为zip object，然后转换为dict，最终使用熊猫获得dataframe.

I am trying to complete a datacamp exercise in which I am required to convert 2 lists into a zip object and then into a dict to finally get a dataframe using pandas.

但是，如果我在列表上使用zip()函数并将其转换为dict然后再转换为数据框，则不会出现任何错误，而是简单的外观完美的数据框. 但是说明说我必须先将压缩的对象转换为列表，然后再将其转换为dict().

However, If I use zip() function over the lists and convert them into a dict and then to a dataframe I get no errors but simple a perfect looking dataframe. But the instructions say that i must convert a zipped object into a list first and then convert it into a dict().

我不明白这对我有什么帮助，因为我每次都会得到相同的输出.即一个数据框.
I am using python3

I dont understand how that helps me because I get the same output each time. i.e a dataframe.
I am using python3

list_keys = ['Country', 'Total']
list_values = [['United States', 'Soviet Union', 'United Kingdom'], [1118, 473, 273]]

import pandas as pd

zipped = list(zip(list_keys,list_values))

# Inspect the list using print()
print(zipped)

# Build a dictionary with the zipped list: data
data = dict(zipped)

# Build and inspect a DataFrame from the dictionary: df
df = pd.DataFrame(data)
print(df)

output:

[('Country', ['United States', 'Soviet Union', 'United Kingdom']), ('Total', [1118, 473, 273])]
          Country  Total
0   United States   1118
1    Soviet Union    473
2  United Kingdom    273

没有list()

zipped = zip(list_keys,list_values)

# Inspect the list using print()
print(zipped)

# Build a dictionary with the zipped list: data
data = dict(zipped)

# Build and inspect a DataFrame from the dictionary: df
df = pd.DataFrame(data)
print(df)

output:

<zip object at 0x10c069648>
          Country  Total
0   United States   1118
1    Soviet Union    473
2  United Kingdom    273

推荐答案

我认为dict(zipped)将zip object或list object转换为dictionary.因此，这里转换为list是多余的.

I think dict(zipped) convert zip object or list object to dictionary. So here convert to list is redundant.

但是如果要从python 3中的zip对象创建DataFrame是有问题的，则需要先转换为tuple的list s:

But if want create DataFrame from zip object in python 3 it is problem, need convert to lists of tuples first:

a = ['United States', 'Soviet Union', 'United Kingdom']
b = [1118, 473, 273]
c = ['Country', 'Total']

zipped = zip(a,b)
print(zipped)
<zip object at 0x000000000DC4E8C8>

df = pd.DataFrame(zipped, columns=c)
print(df)
TypeError: data argument can't be an iterator

---

print(list(zipped))
[('United States', 1118), ('Soviet Union', 473), ('United Kingdom', 273)]

df = pd.DataFrame(list(zipped), columns=c)
print(df)

          Country  Total
0   United States   1118
1    Soviet Union    473
2  United Kingdom    273

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