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我想将以下代码（在 pandas 中运行）转换为在 cuDF 中运行的代码。

来自被操作系列的样本数据.head()被插入到第 3 个代码单元格中的 OG 代码中——应该能够复制/粘贴运行。

熊猫中的原始代码

# both are float columns now
# rawcensustractandblock
s_rawcensustractandblock = df_train['rawcensustractandblock'].apply(lambda x: str(x))

# adjust/set new tract number 
df_train['census_tractnumber'] = s_rawcensustractandblock.str.slice(4,11)

# adjust block number
df_train['block_number'] = s_rawcensustractandblock.str.slice(start=11)
df_train['block_number'] = df_train['block_number'].apply(lambda x: x[:4]+'.'+x[4:]+'0' )
df_train['block_number'] = df_train['block_number'].apply(lambda x: int(round(float(x),0)) )
df_train['block_number'] = df_train['block_number'].apply(lambda x: str(x).ljust(4,'0') )

被操纵的数据

# series of values from df_train.['rawcensustractandblock'].head()
data = pd.Series([60371066.461001, 60590524.222024, 60374638.00300401, 
                  60372963.002002, 60590423.381006])

调整代码以从该示例数据开始

这是使用上面提供的数据而不是整个数据框时代码的外观。

根据尝试转换时遇到的错误，此问题属于系列级别，因此将下面的单元格转换为在 cuDF 中执行应该可以解决问题。

import pandas as pd

# series of values from df_train.['rawcensustractandblock'].head()
data = pd.Series([60371066.461001, 60590524.222024, 60374638.00300401, 
                  60372963.002002, 60590423.381006])

# how the first line looks using the series
s_rawcensustractandblock = data.apply(lambda x: str(x))

# adjust/set new tract number 
census_tractnumber = s_rawcensustractandblock.str.slice(4,11)

# adjust block number
block_number = s_rawcensustractandblock.str.slice(start=11)
block_number = block_number.apply(lambda x: x[:4]+'.'+x[4:]+'0' )
block_number = block_number.apply(lambda x: int(round(float(x),0)) )
block_number = block_number.apply(lambda x: str(x).ljust(4,'0') )

预期变化（输出）

df_train['census_tractnumber'].head()

# out
0    1066.46
1    0524.22
2    4638.00
3    2963.00
4    0423.38
Name: census_tractnumber, dtype: object

df_train['block_number'].head()

0    1001
1    2024
2    3004
3    2002
4    1006
Name: block_number, dtype: object
4

2 回答 2

1

您几乎可以将 cuDF 字符串方法（通过 nvStrings）用于您尝试执行的所有操作。将这些浮点数转换为 cuDF 中的字符串会丢失一些精度（尽管在上面的示例中可能无关紧要），因此对于这个示例，我只是事先进行了转换。如果可能，我建议最初将其创建rawcensustractandblock为字符串列而不是浮点列。

import cudf
import pandas as pd
​
gdata = cudf.from_pandas(pd_data.astype('str'))
​
tractnumber = gdata.str.slice(4,11)
blocknumber = gdata.str.slice(11)
blocknumber = blocknumber.str.slice(0,4).str.cat(blocknumber.str.slice(4), '.')
blocknumber = blocknumber.astype('float').round(0).astype('int')
blocknumber = blocknumber.astype('str').str.ljust(4, '0')
​
tractnumber
0    1066.46
1    0524.22
2    4638.00
3    2963.00
4    0423.38
dtype: object

blocknumber
0    1001
1    2024
2    3004
3    2002
4    1006
dtype: object
于 2019-08-15T19:46:56.337 回答
0

for循环解决方案

熊猫（原代码）

import pandas as pd

# data from df_train.rawcensustractandblock.head()
pd_data = pd.Series([60371066.461001, 60590524.222024, 60374638.00300401, 
                     60372963.002002, 60590423.381006])

# using series instead of dataframe
pd_raw_block = pd_data.apply(lambda x: str(x))

# adjust/set new tract number 
pd_tractnumber = pd_raw_block.str.slice(4,11)

# set/adjust block number
pd_block_number = pd_raw_block.str.slice(11)
pd_block_number = pd_block_number.apply(lambda x: x[:4]+'.'+x[4:]+'0')
pd_block_number = pd_block_number.apply(lambda x: int(round(float(x),0)))
pd_block_number = pd_block_number.apply(lambda x: str(x).ljust(4,'0'))


# print(list(pd_tractnumber))
# print(list(pd_block_number))

cuDF（解决方案代码）

import cudf

# data from df_train.rawcensustractandblock.head()
cudf_data = cudf.Series([60371066.461001, 60590524.222024, 60374638.00300401, 
                         60372963.002002, 60590423.381006])

# using series instead of dataframe
cudf_tractnumber = cudf_data.values_to_string()
# adjust/set new tract number
for i in range(len(cudf_tractnumber)):
  funct = slice(4,11)
  cudf_tractnumber[i] = cudf_tractnumber[i][funct]

# using series instead of dataframe
cudf_block_number = cudf_data.values_to_string()
# set/adjust block number
for i in range(len(cudf_block_number)):
  funct = slice(11, None)
  cudf_block_number[i] = cudf_block_number[i][funct]
  cudf_block_number[i] = cudf_block_number[i][:4]+'.'+cudf_block_number[i][4:]+'0'
  cudf_block_number[i] = int(round(float(cudf_block_number[i]), 0))
  cudf_block_number[i] = str(cudf_block_number[i]).ljust(4,'0')


# print(cudf_tractnumber)
# print(cudf_block_number)
于 2019-08-13T13:22:21.510 回答