在Pandas中将列转换为字符串

我有以下数据帧从一个SQL查询:

(Pdb) pp total_rows
     ColumnID  RespondentCount
0          -1                2
1  3030096843                1
2  3030096845                1

我想这样旋转它:

total_data = total_rows.pivot_table(cols=['ColumnID'])

(Pdb) pp total_data
ColumnID         -1            3030096843   3030096845
RespondentCount            2            1            1

[1 rows x 3 columns]


total_rows.pivot_table(cols=['ColumnID']).to_dict('records')[0]

{3030096843: 1, 3030096845: 1, -1: 2}

但我想确保303列被转换为字符串而不是整数，这样我就得到了这个:

{'3030096843': 1, '3030096845': 1, -1: 2}

当前回答

有四种方法将列转换为字符串

1. astype(str)
df['column_name'] = df['column_name'].astype(str)

2. values.astype(str)
df['column_name'] = df['column_name'].values.astype(str)

3. map(str)
df['column_name'] = df['column_name'].map(str)

4. apply(str)
df['column_name'] = df['column_name'].apply(str)

让我们看看每种类型的性能

#importing libraries
import numpy as np
import pandas as pd
import time

#creating four sample dataframes using dummy data
df1 = pd.DataFrame(np.random.randint(1, 1000, size =(10000000, 1)), columns =['A'])
df2 = pd.DataFrame(np.random.randint(1, 1000, size =(10000000, 1)), columns =['A'])
df3 = pd.DataFrame(np.random.randint(1, 1000, size =(10000000, 1)), columns =['A'])
df4 = pd.DataFrame(np.random.randint(1, 1000, size =(10000000, 1)), columns =['A'])

#applying astype(str)
time1 = time.time()
df1['A'] = df1['A'].astype(str)
print('time taken for astype(str) : ' + str(time.time()-time1) + ' seconds')

#applying values.astype(str)
time2 = time.time()
df2['A'] = df2['A'].values.astype(str)
print('time taken for values.astype(str) : ' + str(time.time()-time2) + ' seconds')

#applying map(str)
time3 = time.time()
df3['A'] = df3['A'].map(str)
print('time taken for map(str) : ' + str(time.time()-time3) + ' seconds')

#applying apply(str)
time4 = time.time()
df4['A'] = df4['A'].apply(str)
print('time taken for apply(str) : ' + str(time.time()-time4) + ' seconds')

输出

time taken for astype(str): 5.472359895706177 seconds
time taken for values.astype(str): 6.5844292640686035 seconds
time taken for map(str): 2.3686647415161133 seconds
time taken for apply(str): 2.39758563041687 seconds

Map (str)和apply(str)与其余两种技术相比需要更少的时间

2021-07-30 18:17:20

其他回答

有四种方法将列转换为字符串

1. astype(str)
df['column_name'] = df['column_name'].astype(str)

2. values.astype(str)
df['column_name'] = df['column_name'].values.astype(str)

3. map(str)
df['column_name'] = df['column_name'].map(str)

4. apply(str)
df['column_name'] = df['column_name'].apply(str)

让我们看看每种类型的性能

#importing libraries
import numpy as np
import pandas as pd
import time

#creating four sample dataframes using dummy data
df1 = pd.DataFrame(np.random.randint(1, 1000, size =(10000000, 1)), columns =['A'])
df2 = pd.DataFrame(np.random.randint(1, 1000, size =(10000000, 1)), columns =['A'])
df3 = pd.DataFrame(np.random.randint(1, 1000, size =(10000000, 1)), columns =['A'])
df4 = pd.DataFrame(np.random.randint(1, 1000, size =(10000000, 1)), columns =['A'])

#applying astype(str)
time1 = time.time()
df1['A'] = df1['A'].astype(str)
print('time taken for astype(str) : ' + str(time.time()-time1) + ' seconds')

#applying values.astype(str)
time2 = time.time()
df2['A'] = df2['A'].values.astype(str)
print('time taken for values.astype(str) : ' + str(time.time()-time2) + ' seconds')

#applying map(str)
time3 = time.time()
df3['A'] = df3['A'].map(str)
print('time taken for map(str) : ' + str(time.time()-time3) + ' seconds')

#applying apply(str)
time4 = time.time()
df4['A'] = df4['A'].apply(str)
print('time taken for apply(str) : ' + str(time.time()-time4) + ' seconds')

输出

time taken for astype(str): 5.472359895706177 seconds
time taken for values.astype(str): 6.5844292640686035 seconds
time taken for map(str): 2.3686647415161133 seconds
time taken for apply(str): 2.39758563041687 seconds

Map (str)和apply(str)与其余两种技术相比需要更少的时间

2021-07-30 18:17:20

这是另一个，特别有用的多列转换为字符串，而不是单一的列:

In [76]: import numpy as np
In [77]: import pandas as pd
In [78]: df = pd.DataFrame({
    ...:     'A': [20, 30.0, np.nan],
    ...:     'B': ["a45a", "a3", "b1"],
    ...:     'C': [10, 5, np.nan]})
    ...: 

In [79]: df.dtypes ## Current datatype
Out[79]: 
A    float64
B     object
C    float64
dtype: object

## Multiple columns string conversion
In [80]: df[["A", "C"]] = df[["A", "C"]].astype(str) 

In [81]: df.dtypes ## Updated datatype after string conversion
Out[81]: 
A    object
B    object
C    object
dtype: object

2017-08-23 21:32:24

熊猫版本:1.3.5

更新后的答案

Df ['colname'] = Df ['colname'].astype(str) =>默认情况下这应该有效。但是如果你在使用astype(str)之前创建了str变量，比如str = "myString"，这将不起作用。在本例中，您可能想使用下面的行。

[‘科尔’]

===========

(注:不正确的旧解释)

Df ['colname'] = Df ['colname'].astype('str') =>将数据帧列转换为字符串类型

等等，等等，等等

2022-10-29 20:07:53

使用.apply()和lambda转换函数在这种情况下也有效:

total_rows['ColumnID'] = total_rows['ColumnID']。Apply (lambda x: str(x))