Python 如何将给定数据集中的值添加到空字典中

下面的代码由我正在参加的Python课程练习中的Datacamp版权所有

我得到了一个csv文件，数据集包含Twitter数据，我必须迭代一列中的条目来构建一个字典，其中键是语言的名称，值是给定语言中的推文数量。生成的代码是正确的和有效的然而，我不能完全理解if-else语句部分中的代码是如何工作的

代码的输出是：

{'en'：97，'et'：1，'und'：2}

我的问题是：我们如何获得上述给定的输出。for循环中的代码内部到底发生了什么，以及if-else

---

谢谢大家!

我在

中为| if | else

添加了一些解释性说明，以根据要求提高对代码的理解

为了便于解释，我将数据集更改为一个最小的示例

作为一般提示：Pandas有一个内置方法（

value\u counts

）可以实现同样的功能

# Import pandas
import pandas as pd 

# Import Twitter data as DataFrame: df
# df = pd.read_csv('tweets.csv') 
df = pd.DataFrame(
    data=[
          'en',  # 1st row
          'en',  # 2nd row
          'und', # 3rd row
          'et',  # 4th row
          'und'  # 5th row
          ],
    columns=['lang']
)

# Initialize an empty dictionary: langs_count
langs_count = {}

# Extract column from DataFrame: col
col = df['lang']

print('before the loop, langs_count is an empty dict')
print(langs_count, '\n')
# Iterate over lang column in DataFrame
for ii, entry in enumerate(col):

    # If the language is in langs_count, add 1 
    if entry in langs_count.keys():
        print(f'{ii}\nif: the key "{col.iloc[ii]}" exists, so adds 1 to value')
        langs_count[entry] += 1
    # Else add the language to langs_count, set the value to 1
    else:
        print(f'{ii}\nelse: the key "{col.iloc[ii]}" does not exist, so create it with value 1')
        langs_count[entry] = 1
    print(langs_count, '\n')

# Print the populated dictionary
# print(langs_count)
#{'en': 97, 'et': 1, 'und': 2}

# the same could be reached through
# without the need of loop or if / else
print('value_counts solution')
df['lang'].value_counts().to_dict()

输出：

---

感谢您的清晰解释和您添加的额外知识。它非常有用！

# Import pandas
import pandas as pd 

# Import Twitter data as DataFrame: df
# df = pd.read_csv('tweets.csv') 
df = pd.DataFrame(
    data=[
          'en',  # 1st row
          'en',  # 2nd row
          'und', # 3rd row
          'et',  # 4th row
          'und'  # 5th row
          ],
    columns=['lang']
)

# Initialize an empty dictionary: langs_count
langs_count = {}

# Extract column from DataFrame: col
col = df['lang']

print('before the loop, langs_count is an empty dict')
print(langs_count, '\n')
# Iterate over lang column in DataFrame
for ii, entry in enumerate(col):

    # If the language is in langs_count, add 1 
    if entry in langs_count.keys():
        print(f'{ii}\nif: the key "{col.iloc[ii]}" exists, so adds 1 to value')
        langs_count[entry] += 1
    # Else add the language to langs_count, set the value to 1
    else:
        print(f'{ii}\nelse: the key "{col.iloc[ii]}" does not exist, so create it with value 1')
        langs_count[entry] = 1
    print(langs_count, '\n')

# Print the populated dictionary
# print(langs_count)
#{'en': 97, 'et': 1, 'und': 2}

# the same could be reached through
# without the need of loop or if / else
print('value_counts solution')
df['lang'].value_counts().to_dict()

"""
before the loop, langs_count is an empty dict
{} 

0
else: the key "en" does not exist, so create it with value 1
{'en': 1} 

1
if: the key "en" exists, so adds 1 to value
{'en': 2} 

2
else: the key "und" does not exist, so create it with value 1
{'en': 2, 'und': 1} 

3
else: the key "et" does not exist, so create it with value 1
{'en': 2, 'und': 1, 'et': 1} 

4
if: the key "und" exists, so adds 1 to value
{'en': 2, 'und': 2, 'et': 1} 

value_counts solution
{'en': 2, 'et': 1, 'und': 2}
"""