在PySpark中经常会使用到dataframe数据形式，本篇博文主要介绍，将list转为dataframe时，遇到的数据类型问题。

有如下一个list：

[(22.31670676205784, 15.00427254361571, 14.274554462639939, -48.011495169271186)]

正常情况下：

#!/usr/bin/python
# -*- coding: utf-8 -*-
from pyspark.sql import Row
from pyspark.ml.linalg import Vectors
import numpy as np
from pyspark.ml.classification import LogisticRegression
from pyspark.sql import SparkSession
from pyspark.sql import SQLContext
import os
from pyspark import SparkContext, SparkConf
from pyspark.sql import HiveContext
from pyspark.mllib.classification import LogisticRegressionWithLBFGS



spark = SparkSession \
    .builder \
    .master("yarn") \
    .appName('create_df_test2') \
    .enableHiveSupport() \
    .getOrCreate()


re = [(22.31670676205784, 15.00427254361571, 14.274554462639939, -48.011495169271186)]
print(re)
print(type(re))

df_re = spark.createDataFrame(re,['r1', 'r2', 'r3', 'r'])

由于re中的数据，其实都是float类型的，直接这样写会报错，如下：

这时需要这样处理：

spark = SparkSession \
    .builder \
    .master("yarn") \
    .appName('create_df_test2') \
    .enableHiveSupport() \
    .getOrCreate()


re = [(22.31670676205784, 15.00427254361571, 14.274554462639939, -48.011495169271186)]
print(re)
print(type(re))

df_re = spark.createDataFrame([(float(tup[0]), float(tup[1]), float(tup[2]), float(tup[3])) for tup in re],
                              ['r1', 'r2', 'r3', 'r'])

这样就可以达到效果了。