如何在 for 循環中使用 'iloc'？

Question

`

def start():
    ob = client.futures_order_book(symbol = 'BTCUSDT')
    ob = pd.DataFrame(ob)
    b = ob['bids']
    a = ob['asks']
    n = 1
    for n in range(500):
        b = b.iloc[n]
        print(b)
        n = n+1 

`

返回錯誤AttributeError: 'list' object has no attribute 'iloc'

它在循環之外工作 (`

n = 1
b = b.iloc[n]
print(b)

`)
作品。 但是我需要從 dataframe 中的 500 個字符串中讀取數據。我該如何解決？

Answer 1

在您的示例中，您將b設置為ob['bids'] ，但是，在您的 for 循環中，您將b設置為b.iloc[n] 。 因此，您應該看到至少打印了一個值。 在循環的第二輪， b不再是ob['bids'] ，因此你得到錯誤。

看來您正在嘗試打印bids列的內容。 您可以通過以下方式輕松做到這一點：

print(list(ob['bids']))

這是一個例子：

代碼：

import pandas as pd
import random

# Creating a dataframe for illustration purposes
ob = pd.DataFrame({"bids":[random.randint(1,30) for i in range(500)],
                   "asks":[random.randint(1,35) for i in range(500)]})

print(list(ob['bids']))

Output：

[28, 18, 11, 20, 29, 25, 11, 13, 18, 13, 7, 27, 13, 2, 27, 17, 8, 24, 10, 3, 13, 6, 15, 3, 2, 21, 21, 2, 22, 17, 26, 1, 13, 23, 17, 17, 2, 23, 6, 27, 21, 23, 7, 10, 25, 3, 20, 4, 3, 5, 21, 8, 21, 19, 13, 6, 1, 28, 4, 27, 7, 19, 26, 27, 20, 25, 27, 10, 9, 8, 3, 1, 6, 24, 28, 2, 15, 9, 23, 25, 23, 13, 15, 29, 3, 6, 5, 6, 1, 12, 10, 10, 6, 8, 17, 3, 14, 28, 9, 26, 4, 21, 3, 2, 27, 10, 28, 23, 23, 9, 5, 7, 28, 25, 8, 5, 2, 10, 5, 11, 23, 8, 12, 2, 15, 9, 16, 28, 8, 21, 12, 4, 20, 21, 14, 22, 24, 25, 20, 12, 28, 11, 19, 25, 27, 5, 17, 8, 15, 12, 22, 2, 19, 4, 24, 23, 25, 5, 27, 25, 28, 22, 3, 21, 16, 9, 5, 14, 3, 20, 15, 21, 8, 16, 15, 29, 30, 22, 9, 21, 19, 20, 30, 4, 16, 26, 22, 28, 12, 2, 23, 9, 10, 18, 18, 11, 6, 4, 23, 4, 19, 15, 4, 30, 30, 2, 13, 3, 14, 13, 27, 17, 22, 15, 26, 22, 4, 5, 13, 7, 4, 16, 18, 4, 15, 25, 4, 26, 14, 10, 3, 26, 29, 9, 22, 24, 4, 22, 5, 23, 29, 6, 25, 24, 16, 4, 26, 9, 10, 25, 16, 23, 8, 28, 16, 26, 25, 24, 17, 13, 13, 2, 18, 4, 23, 2, 5, 21, 2, 17, 19, 4, 3, 4, 29, 19, 20, 23, 23, 27, 28, 2, 2, 23, 17, 18, 3, 25, 30, 29, 9, 8, 13, 24, 19, 5, 7, 20, 19, 15, 18, 1, 2, 22, 23, 22, 5, 29, 19, 14, 27, 12, 3, 24, 16, 29, 17, 10, 29, 20, 26, 14, 18, 25, 23, 7, 22, 21, 7, 19, 12, 10, 9, 17, 2, 4, 3, 2, 13, 25, 20, 18, 30, 20, 21, 11, 9, 21, 18, 29, 4, 7, 23, 28, 18, 17, 15, 30, 3, 19, 4, 2, 19, 5, 27, 19, 22, 12, 14, 17, 9, 8, 20, 17, 20, 9, 26, 20, 18, 6, 12, 10, 8, 4, 6, 14, 20, 10, 2, 22, 17, 2, 28, 8, 5, 4, 4, 26, 9, 21, 27, 12, 27, 12, 6, 11, 29, 27, 29, 22, 27, 10, 24, 22, 22, 8, 30, 13, 10, 22, 23, 26, 14, 6, 6, 20, 14, 4, 6, 27, 12, 9, 14, 9, 29, 8, 14, 30, 2, 28, 24, 14, 15, 6, 17, 18, 25, 15, 7, 6, 28, 30, 30, 10, 15, 24, 15, 27, 4, 20, 12, 15, 18, 1, 11, 24, 21, 24, 20, 18, 17, 10, 22, 22, 23, 23, 1, 11, 30, 15, 21, 4, 23, 8, 11, 29, 25, 5, 30, 19, 1, 4, 6, 25, 25, 23, 4, 6, 11, 17]

筆記：

有點不清楚為什么要在 for 循環中遞增n ，因為這應該自動發生，除非您試圖打印每一秒的值？

在我的示例中，您可以像這樣每隔一個數字打印一次：

print(list(ob['bids'])[::2])

Output：

[28, 11, 29, 11, 18, 7, 13, 27, 8, 10, 13, 15, 2, 21, 22, 26, 13, 17, 2, 6, 21, 7, 25, 20, 3, 21, 21, 13, 1, 4, 7, 26, 20, 27, 9, 3, 6, 28, 15, 23, 23, 15, 3, 5, 1, 10, 6, 17, 14, 9, 4, 3, 27, 28, 23, 5, 28, 8, 2, 5, 23, 12, 15, 16, 8, 12, 20, 14, 24, 20, 28, 19, 27, 17, 15, 22, 19, 24, 25, 27, 28, 3, 16, 5, 3, 15, 8, 15, 30, 9, 19, 30, 16, 22, 12, 23, 10, 18, 6, 23, 19, 4, 30, 13, 14, 27, 22, 26, 4, 13, 4, 18, 15, 4, 14, 3, 29, 22, 4, 5, 29, 25, 16, 26, 10, 16, 8, 16, 25, 17, 13, 18, 23, 5, 2, 19, 3, 29, 20, 23, 28, 2, 17, 3, 30, 9, 13, 19, 7, 19, 18, 2, 23, 5, 19, 27, 3, 16, 17, 29, 26, 18, 23, 22, 7, 12, 9, 2, 3, 13, 20, 30, 21, 9, 18, 4, 23, 18, 15, 3, 4, 19, 27, 22, 14, 9, 20, 20, 26, 18, 12, 8, 6, 20, 2, 17, 28, 5, 4, 9, 27, 27, 6, 29, 29, 27, 24, 22, 30, 10, 23, 14, 6, 14, 6, 12, 14, 29, 14, 2, 24, 15, 17, 25, 7, 28, 30, 15, 15, 4, 12, 18, 11, 21, 20, 17, 22, 23, 1, 30, 21, 23, 11, 25, 30, 1, 6, 25, 4, 11]